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ROBINSON'S    MATHEMATICAL    SERIES, 


ELEMENTS 


OF 


GEOMETRY, 


PLANE  AND   SPHERICAL; 


WITH 


NUMEROUS   PRACTICAL   PROBLEMS, 


BY 

HORATIO  N.  ROBINSON,  LL.D., 

AUTHOR  OP  A  FULL  COURSE^OF  MATHEMATICS. 
REWRITTEN  BY 

I.  F.  QUINBY,  A.M.,  LL.D., 

PROFESSOR  OF  MATHEMATICS  AND  NATURAL  PHILOSOPHY,  UNIVER8rTY  OF  EOCHE8TER ; 
AUTHOR  OF  DIFFERENTIAL  AND  INTEGRAL  CALCULUS. 


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NEW  YORK:  CHICAGO:  • 


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),  according  to  Act  of  Congress,  in  the  year  1868,  by 
DANIEL  W.    FISH,  A.M., 
in  the  Clerk's  Office  of  the  District  Court  of  the  United  States  for  the  Eastern 
District  of  New  York. 


PEEFAOE, 


In  the  preparation  of  this  work,  the  author's  previous 
treatise,  Elements  of  Geometry,  has  formed  the  ground- 
work of  construction.  But  in  adapting  the  work  to  the 
present  advanced  state  of  Mathematical  education  in  our 
best  Institutions,  it  was  found  necessary  so  to  alter  the 
plan,  and  the  arrangement  of  subjects,  as  to  make  this 
essentially  a  new  work.  The  demonstrations  of  proposi- 
tions have  undergone  radical  changes,  many  new  proposi- 
tions have  been  introduced,  and  the  number  of  Practical 
Problems  greatly  increased,  so  that  the  work  is  now  believed 
to  be  as  full  and  complete  as  could  be  desired  in  an  elemen- 
tary treatise. 

In  view  of  the  fact  that  the  Seventh  Book  is  so  much 
larger  than  the  others,  it  may  be  asked  why  it  is  not  divided 
into  two.  We  answer,  that  classifications  and  divisions 
are  based  upon  differences,  and  that  the  differences  seized 
upon  for  this  purpose  must  be  determined  by  the  nature  of 
the  properties  and  relations  we  wish  to  investigate.  There 
is  such  a  close  resemblance  between  the  geometrical  prop- 
erties of  the  polyedrons  and  the  round  bodies,  and  the 
demonstrations  relating  to  the  former  require  such  slight 
modifications  to  become  applicable  to  the  latter,  that  there 
seems  no  sufficient  reason  for  separating  into  two  Books 
that  part  of  Geometry  which  treats  of  them. 

M5498G7 


iT  PREFACE. 

Practical  rules  with  applications  will  be  found  throughout 
the  work,  and  in  addition  to  these,  there  is  a  full  collection 
of  carefully  selected  Practical  Problems.  These  are  given 
to  exercise  the  powers  and  test  the  proficiency  of  the  pupil, 
and  when  he  has  mastered  the  most  or  all  of  them,  it  is 
not  likely  that  he  will  rest  satisfied  with  present  acquisi- 
tion, but,  conscious  of  augmented  strength  and  certain  of 
reward,  he  will  enter  new  fields  of  investigation. 

The  author  has  been  aided,  in  the  preparation  of  the 
present  work,  by  I.  F.  Quinby,  A.M.,  of  the  University  of 
Eochester,  N.  Y.,  late  Professor  of  Mathematics  in  the 
United  States  Military  Academy  at  West  Point.  The 
thorough  scholarship  and  long  and  successful  experience 
of  this  gentleman  in  the  class-room,  eminently  qualify  him 
for  such  a  task;  and  to  him  the  public  are  indebted  for* 
much  that  is  valuable,  both  in  the  matter  and  arrangement 
of  this  treatise. 

October,  1860. 


CONTENTS. 


PLANE   GEOMETRY. 

DEFINITIONS. 

Geometrical  Magnitudes ... ... .......  . .. .  Page  9 

Plane  Angles 10 

Plane  Figures  of  Three  Sides w*.-~.U^ 12 

Plane  Figures  of  Four  Sides. .....-^. ...... ....... 13 

The  Circle ......... ..... ... 14 

Units  of  Measure .................. 15 

Explanation  of  Terms. 16 

Postulates.. —...-...-r. .. . 16 

Axioms -. .. 17 

Abbreviations 17 

BOOK  I. 
Of  Straight  Lines,  Angles,  and  Polygons...  ....... -^.. ... .  19 

BOOK  II. 
Proportion,  and  its  Application  to  Geometrical  Investigations. ...    59 

BOOK  III. 
Of  the  Circle,  and  the  Investigation  of  Theorems  dependent  on  its 

Properties 88 

1*  M 


ri  CONTENTS. 

BOOK  IV. 

Problems  in  the  Construction  of  Figures  in  Plane  Geometry..  ...  Ill 

BOOK  V. 

On  the  Proportionalities  and  Measurement  of  Polygons  and  Circles.  130 
Practical  Problems 142 

BOOK  VI. 

On  the  Intersections  of  Planes,  the  Kelative  Positions  of  Planes, 

and  of  Planes  and  Lines 152 

BOOK  VII. 

Solid  Geometry t 172 

Practical  Problems 229 

BOOK  VIII. 

Practical  Geometry.  —  Application  of  Algebra  to  Geometry,  and 

also  Propositions  for  Original  Investigation 231 

Miscellaneous  Propositions  in  Plane  Geometry 238 

BOOK   IX. 

Spherical  Geometry 243 

Definitions 243 


GEOMETRY. 


DEFINITIONS. 

1.  Geometry  is  the  science  which  treats  of  position,  and 
of  the  forms,  measurements,  mutual  relations,  and  pro- 
perties of  limited  portions  of  space. 

Space  extends  without  limit  in  all  directions,  and  contains  all 
bodies. 

2.  A  Point  is  mere  position,  and  has  no  magnitude. 

3.  Extension  is  a  term  employed  to  denote  that  pro- 
perty of  bodies  by  virtue  of  which  they  occupy  definite 
portions  of  space.  The  dimensions  of  extension  are 
length,  breadth,  and  thickness. 

4.  A  Line  is  that  which  has  extension  in  length  only. 
The  extremities  of  a  line  are  points. 

5.  A  Right  or  Straight  Line  is  one  all  of  whose  parts 
lie  in  the  same  direction. 

6.  A  Curved  Line  is  one  whose  consecutive  parts,  how- 
ever small,  do  not  lie  in  the  same  direction. 

7.  A  Broken  or  Crooked  Line  is 
composed  of  several  straight  lines, 
joined  one  to  another  successively, 
and  extending  in  different  directions. 

When  the  word  line  is  used,  a  straight  line  is  to  be  understood, 
unless  otherwise  expressed. 

8.  A  Surface  or  Superficies  is  that  which  has  extension 
in  length  and  breadth  only. 

9.  A  Plane  Surface,  or  a  Plane,  is  a  surface  such  that 

(9) 


10  GEOMETRY. 

if  any  two  of  its  points  be  joined  by  a  straight  line,  every' 
point  of  this  line  will  lie  in  the  surface. 

10.  A  Curved  Surface  is  one  which  is  neither  a  plane, 
nor  composed  of  plane  surfaces. 

11.  A  Plane  Angle,  or  simply  an  Angle, 
is  the  difference  in  the  direction  of  two 
lines  proceeding  from  the  same  point. 

The  other  angles  treated  of  in  geometry  will  be  named  and  defined 
in  their  proper  connections. 

12.  A  Volume,  Solid,  or  Body,  is  that  which  has  exten- 
sion in  length,  breadth,  and  thickness. 

These  terms  are  used  in  a  sense  purely  abstract,  to  denote  mere 
space  —  whether  occupied  by  matter  or  not,  being  a  question  with 
which  geometry  is  not  concerned. 

Lines,  Surfaces,  Angles,  and  Volumes  constitute  the 
different  kinds  of  quantity  called  geometrical  magnitudes. 

13.  Parallel  Lines  are  lines  which  have  

the  same  direction. 

Hence  parallel  lines  can  never  meet,  however  far  they  may  be 
produced ;  for  two  lines  taking  the  same  direction  cannot  approach 
or  recede  from  each  other. 

Two  parallel  lines  cannot  be  drawn  from  the  same  point;  for  it 
parallel,  they  must  coincide  and  form  one  line. 

PLANE    ANGLES. 

To  make  an  angle  apparent,  the  two 
lines  must  meet  in  a  point,  as  AB  and 
A  0,  which  meet  in  the  point  A , 

Angles  are  measured  by  degrees. 

14.  A  Degree  is  one  of  the  three  hundred  and  sixty 
equal  parts  of  the  space  about  a  point  in  a  plane. 

If,  in  the  above  figure,  we  suppose  A  C  to  coincide  with  AB, 
there  will  be  but  one  line,  and  no  angle ;  but  if  AB  retain  its  posi 
tion,  and  A  G  begin  to  revolve  about  the  point  A,  an  angle  will  be 
formed,  and  its  magnitude  will  be  expressed  by  that  number  of  the 


fl. 


c 


DEFINITIONS.  11 

860  equal  spaces  about  the  point  A,  which  is  contained  between 
AB  and  A  G. 

Angles  are  distinguished  in  respect  to  magnitude  by 
the  terms  Right,  Acute,  and  Obtuse  Angles.  J 

15.  A  Right  Angle  is  that  formed  by  one 
line  meeting  another,  so  as  to  make  equal 
angles  with  that  other. 

The  lines  forming  a  right  angle  are  perpendicular 
to  *ach  other. 

16.  An  Acute  Angle  is  less  than  a  right 
angle. 

17.  An  Obtuse  Angle  is  greater  than 
a  right  angle. 

Obtuse  and  acute  angles  are  also  called 
oblique  angles;  and  lines  which  are  neither  parallel  nor  perpen- 
dicular to  each  other  are  called  oblique  lines. 

18.  The  Vertex  or  Apex  of  an  angle  is  the  point  in  which 
the  including  lines  meet. 

19.  An  angle  is  commonly  designated  by  a  letter  at  its 
vertex;  but  when  two  or  more  angles  have  their  vertices 
at  the  same  point,  they  cannot  be 
thus  distinguished. 

For  example,  when  the  three  lines 
ABy  A  C,  and  AD  meet  in  the  common 
point  Aj  we  designate  either  of  the  an- 
gles formed,  by  three  letters,  placing 
that  at  the  vertex  between  those  at  the 
opposite  extremities  of  the  including 
lines.  Thus,  we  say,  the  angle  BAG, 
etc.  B 

20.  Complements.  —  Two  angles  are  said  to  be  comple 
ments  of  each  other,  when  their  sum  is  equal  to  one  right 
angle. 

21.  Supplements.  —  Two  angles  are  said  to  be  supple- 
ments of  each  other,  when  their  sum  is  equal  to  two  n^ht 
angles. 


r 


12  GEOMETRY. 

PLANE    FIGURES. 

22.  A  Plane  Figure,  in  geometry,  is  a  portion  of  a 
plane  bounded  by  straight  or  curved  lines,  or  by  both 
combined. 

23.  A  Polygon  is  a  plane  figure  bounded  by  straight 
Hues,  called  the  sides  of  the  polygon. 

The  least  number  of  sides  that  can  bound  a  polygon  is 
three,  and  by  the  figure  thus  bounded  all  other  polygons 
are  analyzed. 

FIGURES    OF    THREE    SIDES. 

24.  A  Triangle  is  a  polygon  having  three  sides  and 
three  angles. 

Tri  is  a  Latin  prefix  signifying  three ;  hence  a  Triangle  is  lite  < 
/ally  a  figure  containing  three  angled.  Triangles  are  denominated 
from  the  relations  both  of  their  sides  and  angles. 

25.  A  Scalene  Triangle  is  one  in 
which  no  two  sides  are  equal. 

26.  An  Isosceles  Triangle   is  one  in 
which  two  of  the  sides  are  equal. 


27.  An  Equilateral  Triangle  is  one  in 
arhich  the  three  sides  are  equal. 


28.  A  Right -Angled  Triangle  is  one 
which  has  one  of  the  angles  a  right 
angle. 

29.  An  Obtrse-Angled  Triangle  is  one 
1  aving  an  obtuse  angle. 


DEFINITIONS.  13 


30.  An  Acute- Angled  Triangle  is  one 
in  which  each  angle  is  acute. 


31.  An  Equiangular  Triangle   is   one 
having  its  three  angles  jqual. 


Equiangular  triangles  are  dso  equilateral,  and  vice  versa. 
FIGURES     OF    FOUR    SIDES. 

32.  A  Quadrilateral  is  a  polygon  having  four  sides  and 
four  angles. 

33.  A  Parallelogram  is  a  quadrilateral         7 

which  has  its  opposite  sides  parallel.  /  / 

Parallelograms  are  denominated  from  the  rela-  _• 

tions  both  of  their  sides  and  angles. 

34.  A  Rectangle  is  a  parallelogram  hav- 
ing its  angles  right  angles. 

35.  A  Square  is  an  equilateral  rectangle. 


36.  A   Rhomboid  is  an  oblique-angled  parallelogram. 


37.  A  Rhombus  is  an  equilateral  rhom- 
boid. 


38.  A  Trapezium  is  a  quadrilateral  having 
10  two  sides  parallel. 


39.  A   Trapezoid    is    a   quadrilateral    in       f 
hich  two  opposite  s* J 
i  le  other  two  oblique. 


v/hich  two  opposite  sides  are  parallel,  and    / 


40.  Polygous  bounded  by  a  greater  number  of  sides 
2 


14  GEOMETRY. 

than  four  are  denominated  only  by  the  number  of  sides. 
A  polygon  of  five  sides  is  called  a  Pentagon  •  of  six,  a 
Hexagon ;  of  seven,  a  Heptagon  /  of  eight,  an  Octagon ; 
of  nine,  a  JVonagon,  etc. 

4L  Diagonals  of  a  polygon  are  lines 
joining  the  vertices  of  angles  not  ad- 
jacent. 

42.  The  Perimeter  of  a  polygon  is  its  boundary  consid 
ered  as  a  whole. 

43.  The  Base  of  a  polygon  is  the  side  upon  which  the 
polygon  is  supposed  to  stand. 

44.  The  Altitude  of  a  polygon  is  the  perpendicular 
distance  between  the  base  and  a  side  or  angle  opposite 
the  base. 

45.  Equal  Magnitudes  are  those  which  are  not  only 
equal  in  all  their  parts,  but  which  also,  when  applied  the 
one  to  the  other,  will  coincide  throughout  their  whole 
extent. 

46.  Equivalent  Magnitudes  are  those  which,  though  they 
do  not  admit  of  coincidence  when  applied  the  one  to  the 
other,  still  have  common  measures,  and  are  therefore 
numerically  equal. 

47.  Similar  Figures  have  equal  angles,  and  the  same 
number  of  sides. 

Polygons  may  be  similar  without  being  equal ;  that  is,  the  angles 
and  the  number  of  sides  may  be  equal,  and  the  length  of  the  sides 
and  the  size  of  the  figures  unequal. 

X  THE    CIRCLE. 

48.  A  Circle  is  a  plane  figure  bound- 
ed by  one  uniformly  curved  line,  all  of 
the  points  in  which  are  at  the  same 
distance  from  a  certain  point  within, 
called  the  Center, 

49.  The  Circumference  of  a  circle  is 
the  curved  line  that  bounds  it. 


\ 


DEFINITIONS.  15 

50.  The  Diameter  of  a  circle  is  a  line  passing  througn 
its  center,  and  terminating  at  both  ends  in  the  circum- 
ference. 

51.  The  Radius  of  a  circle  is  a  line  extending  from 
its  center  to  any  point  in  the  circumference.  It  is  one 
half  of  the  diameter.  All  the  diameters  of  a  circle  are 
equal,  as  are  also  all  the  radii. 

52.  An  Arc  of  a  circle  is  any  portion  of  the  circum- 
ference. 

53.  An  angle  having  its  vertex  at  the  center  of  a 
circle  is  measured  by  the  arc  intercepted  by  its  sides. 
Thus,  the  arc  AB  measures  the  angle  A  OB ;  and  in  gen- 
eral, to  compare  different  angles,  we  have  but  to  compare 
the  arcs,  included  by  their  sides,  of  the  equal  circles 
having  their  centers  at  the  vertices  of  the  angles. 

UNITS  OF  MEASURE. 

54.  The  Numerical  Expression  of  a  Magnitude  is  a  number 
expressing  how  many  times  it  contains  a  magnitude  of  the 
Bame  kind,  and  of  known  value,  assumed  as  a  unit. 
For  lines,  the  measuring  unit  is  any  straight  line  of  fixed 
value,  as  an  inch,  a  foot,  a  rod,  etc. ;  and  for  surfaces,  the 
measuring  unit  is  a  square  whose  side  may  be  any  linear 
unit,  as  an  inch,  a  foot,  a  mile,  etc.  The  linear  unit 
being  arbitrary,  the  surface  unit  is  equally  so ;  and  its 
selection  is  determined  by  considerations  of  convenience 
and  propriety. 

For  example,  the  parallelogram  A  BBC  is  mea-    c  D 

sured  by  the  number  of  linear  units  in  CD,  mul- 
tiplied by  the  number  of  linear  units  in  AC  ox 
BD;  the  product  is  the  square  units  in  ABDC. 
For,  conceive  CD  to  be  composed  of  any  number    A  B 

of  equal  parts — say  five — and  each  part  some  unit  of  linear  measure, 
and  AC  composed  of  three  such  units;  from  each  point  of  divi- 
sion on  CD  draw  lines  parallel  to  A  C,  and  from  each  point  of  divi- 
sion on  A  C  draw  lines  parallel  to  CD  or  AB ;  then  it  is  as  obvious 


16  GEOMETRY. 

as  an  axiom  that  the  parallelogram  will  contain  5  X  3  =  15  square 
units.  Hence,  to  find  the  areas  of  right-angled  parallelograms,  mul- 
tiply the  base  by  the  altitude. 


EXPLANATION  OF  TERMS. 

55.  An  Axiom  is  a  self-evident  truth,  not  only  too  sim- 
ple to  require,  but  too  simple  to  admit  of,  demonstration. 

56.  A  Proposition  is  something  which  is  either  pro- 
posed to  be  done,  or  to  be  demonstrated,  and  is  either  a 
problem  or  a  theorem. 

57.  A  Problem  is  something  proposed  to  be  done. 

58.  A  Theorem  is  something  proposed  to  be  demon- 
strated. 

59.  A  Hypothesis  is  a  supposition  made  with  a  view  to 
draw  from  it  some  consequence  which  establishes  the 
truth  or  falsehood  of  a  proposition,  or  solves  a  problem. 

60.  A  Lemma  is  something  which  is  premised,  or  demon- 
strated, in  order  to  render  what  follows  more  easy. 

61.  A  Corollary  is  a  consequent  truth  derived  imme- 
diately from  some  preceding  truth  or  demonstration. 

62.  A  Scholium  is  a  remark  or  observation  made  upon 
something  going  before  it. 

63.  A  Postulate  is  a  problem,  the  solution  of  which  is 
self-evident. 

POSTULATES. 

Let  it  be  granted — 

I.  That  a  straight  line  can  be  drawn  from  any  one  poirt 
to  any  other  point ; 

IL.  That  a  straight  line  can  be  produced  to  any  distance, 
or  terminated  at  any  point ; 

III.  That  the  circumference  of  a  circle  can  be  de- 
Bcrjbed  about  any  center,  at  any  distance  from  that  center. 


DEFINITIONS.  17 


AXIOMS. 


1.  Things  which  are  equal  to  the  same  thing  are  equal  U 
each  other. 

2.  Wlien  equals  are  added  to  equals  the  wholes  are  equal, 

3.  When  equals  are  taken  from  equals  the  remainders  are 
equal. 

4.  When  equals  are  added  to  unequals  the  wholes  are 
unequal. 

5.  Wlien  equals  are  taken  from  unequals  the  remainders 
are  unequal. 

6.  Things  which  are  double  of  the  same  thing,  or  equal 
things,  are  equal  to  each  other. 

7.  Things  which  are  halves  of  the  same  thing,  or  of  equal 
things,  are  equal  to  each  other. 

8.  The  whole  is  greater  than  any  of  its  parts. 

9.  Every  whole  is  equal  to  all  its  parts  taken  together. 

10.  Things  which  coincide,  or  fill  the  same  space,  are 
identical,  or  mutually  equal  in  all  their  parts. 

11.  All  right  angles  are  equal  to  one  another. 

12.  A  straight  line  is  the  shortest  distance  between  two 
points. 

18.  Two  straight  lines  cannot  inclose  a  space. 

ABBREVIATIONS. 

The  common  algebraic  signs  are  used  in  this  work, 
and  demonstrations  are  sometimes  made  through  the 
medium  of  equations ;  and  it  is  so  necessary  that  the 
student  in  geometry  should  understand  some  of  the  more 
simple  operations  of  algebra,  that  we  assume  that  he  is 
acquainted  with  the  use  of  the  signs.  As  the  terms 
circle,  angle,  triangle,  hypothesis,  axiom,  theorem,  cor- 
ollary, and  definition,  are  constantly  occurring  in  a  course 
of  geometiy,  we  shall  abbreviate  them  as  shown  in  the 
following  list : 


20  GEOMETRY. 

By  Th.  1,  any  two  supplementary  angles,  as  ABD^ 
ABO,  are  together  equal  to  two  right  angles.  And  since 
the  angular  space  about  the  point  B  is  neither  increased 
nor  diminished  by  the  number  of  lines  drawn  from  that 
point,  the  sum  of  all  the  angles  DBA,  ABU,  JEBH, 
HBQ,  fills  the  same  spaces  as  any  two  angles  HBD, 
HBO.  Hence  the  theorem  ;  from  any  point  in  a  line,  the 
sum  of  all  the  angles  that  can  be  formed  on  the  same  side  of 
the  line  is  equal  to  two  right  angles. 

Cor.  1.  And,  as  the  sum  of  all  the  angles  that  can  be 
formed  on  the  other  side  of  the  line,  OB,  is  also  equal  to 
two  right  angles;  therefore,  all  the  angles  that  can  be 
formed  quite  round  a  point,  B,  by  any  number  of  lines,  are 
together  equal  to  four  right  angles. 

Oor.  2.  Hence,  also,  the  whole  circum- 
ference of  a  circle,  being  the  sum  of  the 
measures  of  all  the  angles  that  can  be 
made  about  the  center  F,  (Def.  53),  is  the 
measure  of  four  right  angles;  conse- 
quently, a  semicircumference.  is  the  mea- 
sure of  two  right  angles ;  and  a  quadrant,  or  90°,  is  the 
measure  of  one  right  angle. 


THEOREM   III. 


X 

If  one  straight  line  meets  two  other  straight  lines  at  a 
common  point,  forming  two  angles,  which  together  are  equal 
tc  two  right  angles  the  'two  straight  lines  are  one  and  the. 
tame  line. 

Let  the  line  AB  meet  the 
lines  BD  and  BE  at  the  com- 
mon point  B,  making  the  sum 
of  tne  two  angles  ABB,  ABB, 
equal  to  two  right  angles;  We  j£ 
are  to  prove  that  DB  and  BE* 


are  one  straight  line 


j^' 


BOOK    I.  21 

If  I)B  and  BE  are  not  in  the  same  line,  produce  DB 
to  0,  thus  forming  one  line,  BBC 

Now  by  Th.  1,  ABB  +  ABO  must  be  equal  to  two 
right  angles.  But  by  hypothesis,  ABB  4-  ABE  is  equal 
to  two  right  angles. 

*  Therefore,  ABB  +  ABO  is  equal  to  ABB  -f  ABE, 
(Ax.  1).  From  each  of  these  equals  take  away  the  com- 
mon angle  ABB,  and  the  angle  ABO  will  be  equal  to 
ABE,  (Ax.  3).  That  is,  the  line  BE  must  coincide  with 
BO,  and  they  will  be  in  fact  one  and  the  same  line,  and 
they  cannot  be  separated  as  is  represented  in  the  figure. 

Hence  the  theorem ;  if  one  line  meets  two  other  lines  at  a 
common  point,  forming  two  angles  which  together  are  equal 
to  two  right  angles,  the  two  lines  are  one  and  the  same  line. 

THEOKEM    IV. 

If  two  straight  lines  intersect  each  other,  the  opposite  or 
vertical  angles  must  be  equal. 

If  AB  and  OB  intersect  each 
other  at  E,  we  are  to  demonstrate 
that  the  angle  AEO  is  equal  to 
the  vertical  angle  BEB ;  and  the 
angle  AEB,  to  the  vertical  angle 
OEB. 

As  AB  is  one  line  met  by  BE,  another  line,  the  two 
angles  AEB  and  BEB,  on  the  same  side  of  AB,  are  equal 
to  two  right  angles,  (Th.  1).  Also,  because  OB  is  a  right 
line,  and  AE  meets  it,  the  two  angles  AEO  and  AEB 
are  together  equal  to  two  right  angles. 

Therefore,  AEB  +  BEB  =  AEO  +  AEB.     (Ax.  1.) 

If  from  these  equals  we  take  away  the  common  angle 
AEB,  the  remaining  angle  BEB  must  be  equal  to  the 
remaining  angle  AEO,  (Ax.  3).  In  like  manner,  we  can 
prove  that  AEB  is  equal  to  OEB.  Hence  the  theo**en. ; 
if  the  two  lines  intersect  each  other,  the  vertic  %l  angle*  mu  it 
be  equal. 


20  GEOMETRY. 

By  Th.  1,  any  two  supplementary  angles,  as  ABD^ 
ABC,  are  together  equal  to  two  right  angles.  And  since 
the  angular  space  about  the  point  B  is  neither  increased 
nor  diminished  by  the  number  of  lines  drawn  from  that 
point,  the  sum  of  all  the  angles  DBA,  ABU,  EBH, 
HBO,  fills  the  same  spaces  as  any  two  angles  HBD, 
HBQ.  Hence  the  theorem ;  from  any  point  in  a  line,  the 
sum  of  all  the  angles  that  can  be  formed  on  the  same  side  of 
the  line  is  equal  to  two  right  angles. 

Cor.  1.  And,  as  the  sum  of  all  the  angles  that  can  be 
formed  on  the  other  side  of  the  line,  CD,  is  also  equal  to 
two  right  angles;  therefore,  all  the  angles  that  can  be 
formed  quite  round  a  point,  B,  by  any  number  of  lines,  are 
together  equal  to  four  right  angles. 

Cor.  2.  Hence,  also,  the  whole  circum- 
ference of  a  circle,  being  the  sum  of  the 
measures  of  all  the  angles  that  can  be 
made  about  the  center  F,  (Def.  53),  is  the 
measure  of  four  right  angles;  conse- 
quently, a  semicircumference,  is  the  mea- 
sure of  two  right  angles ;  and  a  quadrant,  or  90°,  is  the 
measure  of  one  right  angle. 


THEOREM    III. 


t 

If  one  straight  line  meets  two  other  straight  lines  at  a 
common  point,  forming  two  angles,  which  together  are  equal 
tc  two  right  angles  the  'two  straight  lines  are  one  and  the 
tame  line. 

Let  the  line  AB  meet  the 
lines  BD  and  BE  at  the  com-  A 

mon  point  B,  making  the  sum 
of  tne  two  angles  ABB,  ABE, 
equal  to  two  right  angles ;  We 
are  to  prove  that  DB  and  BlM 
are  one  straight  line. 


BOOK    I.  21 

If  1)B  and  BE  are  not  in  the  same  line,  produce  BB 
to  0,  thus  forming  one  line,  BBC. 

Now  by  Th.  1,  ABB  -f  ABO  must  be  equal  to  two 
right  angles.  But  by  hypothesis,  ABB  +  ABB  is  equal 
to  two  right  angles. 

*  Therefore,  ABB  +  ABO  is  equal  to  ABB  +  ABB, 
(Ax.  1).  From  each  of  these  equals  take  away  the  com- 
mon angle  ABB,  and  the  angle  ABO  will  be  equal  to 
ABB,  (Ax.  3).  That  is,  the  line  BE  must  coincide  with 
BO,  and  they  will  be  in  fact  one  and  the  same  line,  and 
they  cannot  be  separated  as  is  represented  in  the  figure. 

Hence  the  theorem ;  if  one  line  meets  two  other  lines  at  a 
common  point,  forming  two  angles  which  together  are  equal 
to  two  right  angles,  the  two  lines  are  one  and  the  same  line. 

THEOREM    IV. 

If  two  straight  lines  intersect  each  other,  the  opposite  or 
vertical  angles  must  be  equal. 

If  AB  and  OB  intersect  each 
other  at  E,  we  are  to  demonstrate 
that  the  angle  AEO  is  equal  to 
the  vertical  angle  BEB ;  and  the 
angle  AEB,  to  the  vertical  angle 
OEB. 

As  AB  is  one  line  met  by  BE,  another  line,  the  two 
angles  AEB  and  BEB,  on  the  same  side  of  AB,  are  equal 
to  two  right  angles,  (Th.  1).  Also,  because  OB  is  a  right 
line,  and  AE  meets  it,  the  two  angles  AEO  and  AEB 
are  together  equal  to  two  right  angles. 

Therefore,  AEB  +  BEB  =  AEO  +  AEB.     (Ax.  1.) 

If  from  these  equals  we  take  away  the  common  angle 
AEB,  the  remaining  angle  BEB  must  be  equal  to  the 
remaining  angle  AEO,  (Ax.  3).  In  like  manner,  we  can 
prove  that  AEB  is  equal  to  OEB.  Hence  the  thecen. ; 
if  the  two  lines  intersect  each  other,  the  vertic  il  angle*  km  it 
be  equal. 


22  GEOMETRY. 

Second  Demonstration. 

By  Def.  11,  the  angle  DEB  is  the  difference  in  the 
direction  of  the  lines  ED  and  EB ;  and  the  angle  AEQ 
is  the  difference  in  the  direction  of  the  lines  EC  and  EA. 

But  ED  is  opposite  in  direction  to  EC;  and  EB  is 
opposite  in  direction  to  EA. 

Hence,  the  difference  in  the  direction  of  ED  and  EB 
is  the  same  as  that  of  EC  and  EA>  as  is  obvious  by  in- 
spection. 

Therefore,  the  angle  DEB  is  equal  to  its  opposite  AEC 

In  like  manner,  we  may  prove  AED  =  CEB. 

Hence  the  theorem ;  if  two  lines  intersect  each  other,  ths 
vertical  angles  must  be  equal. 

THEOREM   V. 

If  a  straight  line  intersects  two  parallel  lines,  the  sum  of 
the  two  interior  angles  on  the  same  side  of  the  intersecting 
line  is  equal  to  two  right  angles. 

[Note.  —  By  interior  angles,  we  mean  angles  which  lie  between  the 
parallels ;  the  exterior  angles  are  those  not  between  the  parallels.] 

Let  the  line  EF  intersect  the 
parallels  AB  and  CD  \  then 
we  are  to  demonstrate  that 
the  angles  BGE  +  GHD  =        

2K.L  

Because  GB  and  ED  are       C  /H  d 

parallel,  they  are  equally  in-  <£ 

clined  to  the  line  EF,  or  have 

the  same  difference  of  direction  from  that  line.  There- 
fore, |_  FGB  =  [__  GHD.  To  each  of  these  equals  add 
the  \_BGH,  and  we  have  FGB  +BGH=GITD+BGB:. 
But  by  Th.  1,  the  first  member  of  this  equation  is  equal 
to  two  right  angles ;  and  the  second  member  is  the  sum 
of  the  two  angles  between  the  parallels.  Hence  the  theo- 
rem ;  if  a  line  intersects  two  parallel  lines,  the  sum  of  the  two 
interior  angles  on  the  same  side  of  the  intersecting  line  must 
be  equal  to  two  right  angle*. 


BOOK    I.  23 

Sctt&Ln*  —  As  AB  and  CD  are  parallel  lines,  and  EF  is  a  line 
intersecting  them,  AB  and  EF  must  make  angles  equal  to  those  made 
by  CD  and  EF.  That  is,  the  angles  about  the  point  G  must  be  equal 
to  the  corresponding  angles  about  the  point  H. 

THEOREM    VI. 

If  a  line  intersects  two  parallel  lines,  the  alternate  interior 
angles  are  equal. 

Let  AB  and  CD  be  paral- 
lels, intersected  by  EF  at  E 
and  G.   Then  we  are  to  prove 

that  the  angle  A  GE  is  equal        

to  the  alternate  angle  GED, 

and  QEG  =  EGB.  c  ~7W  "  D 

By  Th.  5,  [_BGE  +  \__  / 

GED  =  two  right  angles.  Al- 
so, by  Th.  1,  \_AGE  +  [_BGE  =  two  right  angles. 
From  these  equals  take  away  the  common  angle  BGE, 
and  L  GHD  will  be  left,  equal  to  \_AGE,  (Ax.  3).  In 
like  manner,  we  can  prove  that  the  angle  QEG  is  equal 
to  the  angle  EGB.  Hence  the  theorem ;  if  a  line  intersects 
two  parallel  lines,  the  alternate  interior  angles  are  equal. 

Cor.  1.  Since        [__  A  GE  =  [__  FGB, 

and  \_AGE=[__GED; 

Therefore,  L  FGB  -  |_  &HD  (Ax.  1). 

Also,  L  A  &F  +  L  A  GH  =  2  R-  L>  (Th.  1), 

and  L  OEG  +  [_AGE  =  2  K.  L,  (Th.  5); 

Therefore 

\_AGF  +  l_AGE=  [_CEG  +  L_AGE,(Ax.l); 

and  L  A  GF  =  L  OEG,  (Ax.  3). 

That  is,  the  exterior  angle  is  equal  to  the  interior  opposiU 
angle  on  the  same  side  of  the  intersecting  line. 

Cor.  2.  Since      \_AGE  =  [_FGB, 

and  l_AGE=\_CEE; 

Therefore,  [__FGB  =  |_  OEE. 

In  the  same  manner  it  may  be  shown  that 
[__AGF  =  \_EED. 

Hence,  the  alternate  exterior  angles  are  equal. 

v  * 


GEOMETKY. 


THEOREM    VII. 


If  a  line  intersects  two  other  lines,  making  the  sum  of  the 
two  interior  angles  on  the  same  side  of  the  intersecting  line 
equal  to  two  right  angles,  the  two  straight  lines  are  parallel. 

Let  the  line  EF  intersect 
the  lines  AB  and  OB,  making 
the  two  angles  BQR  +  GHB  A 
=  to  two  right  angles ;  then 
we  are  to  demonstrate  that 
AB  and  OB  are  parallel.  C  /H  "~ d 

As  EF  is  a  right  line  and  E' 

BGr  meets  it,  the  two  angles 

FCrB  and  BGH  are  together  equal  to  two  right  angles, 
(Th.  1).  But  by  hypothesis,  the  angles,  BGHnnd  GHB, 
are  together  equal  to  two  right  angles.  From  these  two 
equals  take  away  the  common  angle  BGrH,  and  the  re- 
maining angles  FGB  and  GHD  must  be  equal,  (Ax.  3). 
Now,  because  GB  and  HD  make  equal  angles  with  the 
same  line  EF,  they  must  extend  in  the  same  direction ;  and 
lines  having  the  same  direction  are  parallel,  (Def.  13). 
Hence  the  theorem ;  if  a  line  intersects  two  other  lines,  making 
the  sum  of  the  two  interior  angles  on  the  same  side  of  the  in- 
tersecting line  equal  to  two  right  angles,  the  two  lines  must  b« 
'parallel. 

Cor.  1.  If  a  line  intersects  two  other  lines,  making  the 
alternate  interior  angles  equal,  the  two  lines  intersected 
must  be  parallel. 

Suppose  the  \_  AGH  =  |__  &HI)-  Adding  L  H&B 
to  each,  we  have 

[_AGH  +  [_HGB  =  L_  GHB  +  [__HGB. 
but  the  first  member  of  this  equation,  that  is,  [_AGfH-+ 
|__  HGB,  is  equal  to  two  right  angles ;  hence  the  second 
member  is  also  equal  to  the  same ;  and  by  the  theorem, 
the  lines  AB  and  OB  are  parallel. 

Cor.  2.  If  a  line  intersects  two  other  lines,  making  the 


BOOK    I.  25 

opposite  exterior  and  interior  angles  equal,  the  two  lines 

intersected  must  be  parallel. 
Suppose  the  [_  FGB  =  [_  GRD.    Adding  the  [__  HGB 

to  each,  we  have 

[_FGB  +  \_EGB  =  L  &HD  +  ##£. 

But  the  first  member  of  this  equation  is  equal  to  two 

right  angles ;  hence  the  second  member  is  also  equal  to 

two  right  angles ;  and  by  the  theorem,  the  lines  AB  and 

OB  are  parallel. 

Oor.  3.  If  a  line  intersects  two  other  lines,  making  tbo 

alternate  exterior  angles  equal,  the  lines  must  be  parallel: 
Suppose     [_BGF=[_ORF,  and  [_AGF  =  [_DRF, 
ByTh.4,    [_BGF^[__AGR,2ind[_ORF  =  [_DRG 
And  since  [_BGF  =  [_ORF,         [__AGR=[_DRG . 
That  is,  the  alternate  interior  angles  are  equal;  an* J 

hence  (by  Cor.  1)  the  two  lines  are  parallel. 

THEOREM    VIII. 

If  two  angles  have  their  sides  parallel,  the  two  angles  will 
be  either  equal  or  supplementary. 

Let  A  0  be  parallel  to  BD,  and  AH 
parallel  to  BF  or  to  BG.  Then  we  are 
to  pror^  that  the  angle  DBx  is  equal 
to  the  angle  OAR,  and  that  the  angle 
DBG  is  supplementary  to  the  angle  A. 
The  angle  OAR  is  formed  by  the  differ- 
ence in  the  direction  of  A  0  and  A  R;  and 
the  angle  DBF  is  formed  by  the  differ- 
ence in  the  direction  of  BB  and  BF. 
But  A  0  and  A  R  have  the  same  direc- 
tions as  BB  and  BF,  because  they  are  respectively  paral- 
lel. Therefore,  by  Def.  11,  L  CAR=  \__BBF.  But  the 
line  BG  has  the  same,  direction  as  BF,  and  the  angle 
DBG  is  supplementary  to  DBF.  Hence  the  theorem; 
angles  whose  sides  are  parallel  are  either  equal  or  supple* 
inentary. 
3 


X 


\ 


26  GEOMETRY. 

v,  •  THEOREM    IX. 

The  opposite  angles  of  any  parallelogram  we  equal. 

Let  AEBG  be  a  parallel- 
ogram.     Then   we    are   to  \ 

prove  that  the  angle  GBE     6  \] 

is  equal  to  its  opposite  angle 
A. 

Produce  EB  to  D,  and  GB 
to  F;  then,  since  BD  is  par- 
allel to  AG,  and  BF  to  AE,  the  angle  DBF  is  equal  co 
the  angle  A,  (Th.  8). 

But  the  angles  GBE  and  DBF,  being  vertical,  are 
equal,  (Th.  4).  Therefore,  the  opposite  angles  GBE  and 
A,  of  the  parallelogram  AEBG,  are  equal. 

In  like  manner,  we  can  prove  the  angle  E  equal  to 
the  angle  G.  Hence  the  theorem ;  the  opposite  angles  of 
any  parallelogram  are  equal. 

THEOREM    X. 

The  sum  of  the  angles  of  any  parallelogram  i»  equal  to 
four  right  angles. 

Let  ABQD  be  a  parallelo- 
gram.  We  are  to  prove  that 
the  sum  of  the  angles  A,  B,  0 
and  D,  is  equal  to  four  right 
angles,  or  to  360°. 

Because  AD  and  BO  are  parallel  lines,  and  AB  inter- 
sects them,  the  two  interior  angles  A  and  B  are  together 
equal  to  two  right  angles,  (Th.  5).  And  because  CD  in- 
tersects the  same  parallels,  the  two  interior  angles  O  and 
D  are  also  together  equal  to  two  right  angles.  By  addi- 
tion, we  have  the  sum  of  the  four  interior  angles  of  the 
parallelogram  ABCD,  equal  to  four  right  angles.  Hence 
the  theorem ;  the  sum  of  the  angles  of  any  parallelogram  is 
taual  tc  four  right  angles. 

is.  x 


A 


BOOK    I. 


THEOREM    XI. 


27 

V 


The  sum  of  the  three  angles  of  any  triangle  is  equal  to  two 
right  angles. 

Let  A B  0  be  a  triangle, 
and  through  its  vertex  0 
draw  a  line  parallel  to  the 
base  AB,  and  produce 
the  sides  A C  and  BO. 
Then  the  angles  A  and 
a,  being  exterior  and  in- 
terior opposite  angles  on 
the  same  side  of  the  line  AC,  are  equal  to  each  other. 
For  the  same  reason,  [_  E  =  L  b.  The  angles  C  and  c, 
being  vertical  angles,  are  also  equal,  (Th.  4).  Therefore, 
the  angles  A,  B,  C  are  equal  to  the  angles  a,  b,  c  respect- 
ively. But  the  angles  around  the  point  C,  on  the  upper 
side  of  the  parallel  CD,  are  equal  to  two  right  angles, 
(by  Th.  2).  Hence  the  theorem ;  the  sum  of  the  three 
angles,  etc. 

Second  Demonstration. 
Let  AEBG-  be  a  parallelogram 


Draw  the  diagonal  GE ;  thus  di- 
viding the  parallelogram  into  two 
triangles,  and  the  opposite  angles 
G  and  E  each  into  two  angles. 

Because  GB  and  AE  are  parallel,  the  alternate  interior 
angles  BGE  and  GEA  are  equal,  (Th.  6).  Designate 
each  of  these  by  b. 

In  like  manner,  because  EB  and  AG  are  parallel,  the 
alternate  interior  angles,  BEG  and  EGA,  are  equal. 
Designate  each  of  these  by  a. 

Now  we  are  to  prove  that  the  three  angles  B,  b,  and  a, 
and  also  that  the  three  angles  A,  a,  and  b,  are  equal  to 
two  right  angles. 


28  GEOMETRY. 

Because  A  and  B  are  opposite  angles  of  a  parallelo- 
gram, they  are  equal,  (Th.   9),  and  [_A  -f  [_B  =  2  [__A. 

And  all  the  interior  angles  of  the  parallelogram  are 
equal  to  four  right  angles,  (Th.  10). 

Therefore,  2 A  +  2a  -f-  2b  =  4  right  angles. 

Dividing  by  2,  and  J.  +    a  +    &  =  2  " 

That  is,  all  the  angles  of  the  triangle  AGrE  are  together 
equal  to  two  right  angles 

Hence  the  theorem  ;  the  sum  of  the  three  angles,  etc. 

Scholium. — Any  triangle,  as  AGE,  may  be  conceived  to  be  part  of 
a  parallelogram.  For,  let  A  GE  be  drawn  independently  of  the  paral- 
lelogram ;  then  draw  EB  from  the  point  E  parallel  to  A  G,  and  through 
the  point  G  draw  GB  parallel  to  AE,  and  a  parallelogram  will  be 
formed  embracing  the  triangle  ;  and  thus  the  sum  of  the  three  angles 
of  any  triangle  is  proved  equal  to  two  right  angles. 

This  truth  is  so  fundamental,  important,  and  practical, 
as  to  require  special  attention ;  we  therefore  give  a 

Third  Demonstration. 

Let  ABQ  be  a  triangle.  Then 
we  are  to  show  that  the  angles  A, 
Q,  and  ABQ,  are  together  equal 
*o  two  right  angles. 

Let  AB  be  produced  to  D,  and 
from  B  draw  BE  parallel  to  A  0. 

Then,  EBB  and  CAB  being  exterior  and  interior  op- 
posite angles  on  the  same  side  of  the  line  AB,  are  equal, 
(Th.  6,  Cor.1).  Also,  QBE  and  AQB,  being  alternate 
angles,  are  equal,  (Th.  6). 

By  addition,  observing  that  [__  QBE,  added  to  \__EBB, 
must  make  |__  QBD,  we  have 

l_CBB  =  l_A  +  \_Q.  (1.) 

To  each  of  these  equals  add  the  angle  QBA,  and  we 
shall  have 

L  QBA  +  L  OBB  =L^  +  L_tf+L  OB  A. 

But  (by  Th.  1),  the  sum  of  the  first  two  is  equal  to  two 


BOOK    I.  29 

rigltt  angles;  therefore,  the  three  angles,  A,  C,  and  CBAf 
are  together  equal  to  two  right  angles. 

Hence  the  theorem ;  the  sum  of  the  three  angles,  etc. 

y  THEOREM    XII. 

If  any  side  of  a  triangle  is  produced,  the  exterior  angle  i* 
equal  to  the  sum  of  the  two  interior  opposite  angles. 

Let  ABO  be  a  triangle.  Pro- 
duce AB  to  B;  and  we  are  to 
prove  that  the  angle  CBB  is  equal 
to  the  sum  of  the  two  angles  A 
and  C. 

We  establish  this  theorem  by  a 
course  of  reasoning  in  all  respects  the  same  as  that  hy 
which  we  obtained  Eq.  (1.),  third  demonstration,  (Th.  11). 

Cor.  1.  Since  the  exterior  angle  of  any  triangle  is  equal 
to  the  sum  of  the  two  interior  opposite  angles,  therefore 
it  is  greater  than  either  one  of  them. 

Cor.  2.  If  two  angles  in  one  triangle  be  equal  to  two 
angles  in  another  triangle,  the  third  angles  will  also  be 
equal,  each  to  each,  (Ax.  3);  that  is,  the  two  triangles 
will  be  mutually  equiangular. 

Cor.  3.  If  one  angle  in  a  triangle  be  equal  to  one  angle 
in  another,  the  sum  of  the  remaining  angles  in  the  one 
will  also  be  equal  to  the  sum  of  the  remaining  angles  in 
the  other,  (Ax.  3). 

Cor.  4.  If  one  angle  of  a  triangle  be  a  right'  angle,  the 
sum  of  the  other  two  will  be  equal  to  a  right  angle, 
and  each  of  them  singly  will  be  acute,  or  less  than  a  right    -. 
angle. 

Cor.  5.  The  two  smaller  angles  of  every  triangle  are 
acute,  or  each  is  less  than  a  right  angle. 

Cor.  6.  All  the  angles  of  a  triangle  may  be  acute,  but 
no  triangl  3  can  have  more  than  one  right  or  one  obtuse 
angle. 


30  GEOMETRY. 

V  THEOREM    XIII. 

In  any  polygon,  the  sum  of  all  the  interior  angles  is  equal 
to  twice  as  many  riglit  angles,  less  four,  as  the  figure  has 


Let  ABODE  be  any  polygon  ; 
we  are  to  prove  that  the  sum  of 
all  its  interior  angles,  A+B  +  O 
+  D+E,  is  equal  to  twice  as 
many  right  angles,  less  four,  as 
the  figure  has  sides. 

From  any  point,  p,  within  the 
figure,  draw  lines  pA,  pB,  pG,  etc.,  to  all  the  angles, 
thus  dividing  the  polygon  into  as  many  triangles  as  it 
has  sides.  Now,  the  sum  of  the  three  angles  of  each  of 
these  triangles  is  equal  to  two  right  angles,  (Th.  11)  j  and 
the  sum  of  the  angles  of  all  the  triangles  must  be  equal 
to  twice  as  many  right  angles  as  the  figure  has  sides.  But 
the  sum  of  these  angles  contains  the  sum  of  four  right 
angles  about  the  point  p  ;  taking  these  away,  and  the 
remainder  is  the  sum  of  the  interior  angles  of  the  figure. 
Therefore,  the  sum  must  be  equal  to  twice  as  many  right 
angles,  less  four,  as  the  figure  has  sides. 

Hence  the  theorem  ;  in  any  polygon,  etc. 

From  this  Theorem  is  derived  the  rule  for  finding  the 
sum  of  the  interior  angles  of  any  right-lined  figure  : 

Subtract  2  from  the  number  of  sides,  and  multiply  the  re- 
mainder by  2 ;  the  product  will  be  the  number  of  right  angles. 

Thus,  if  the  number  of  sides  be  represented  by  S,  the 
number  of  right  angles  will  be  represented  by  (2#— 4). 

The  Theorem  is  not  varied  in  case  n 

of   a    re-entrant    angle,    as    repre-  ^/"^ 

sen  ted  at  d,  in  the  figure  ABCdEF.     -^^^^    L 

Draw  lines  from  the  angle  d  to       \  "/( 7E 

the  several  opposite  angles,  making        \      /     \       / 

as  many  triangles  as  the  figure  has         \/ \y 

sides,  less  two,  and  the  sum  of  the        A  F 

three  angles  of  each  triangle  equals  two  right  angles. 


X 


BOOK    I.  31 

THEOREM    XIV. 

If  the  sides  of  one  angle  be  respectively  perpendicular 
to  the  sides  of  a  second  angle,  these  two  angles  will  be  either 
equal  or  supplementary. 

Let  BAD  be  the  first  angle,  and 
from  any  point  within  it,  as  C,  draw  / 

CB  and  CD,  at   right   angles,  the         3T^>C 
first  to  AB}  and  the  second  to  AD,  I        \    ^""-Jl 

and  produce   CD  in  the  direction        /  j 

CD,  thus  forming  at  C  the  supple-      A  B 

entary  angles  BCE,  BCD  ;  then 
will  the  angle  BCD  be  equal  to  the  angle  A,  and  therefore 
BCD,  which  is  the  supplement  of  BCD,  will  also  be  the 
supplement  of  the  angle  A. 

For  since  ABCD  is  a  quadrilateral,  the  sum  of  the  four 
interior  angles  is  four  right  angles  (Prop.  13),  and  because 
the  angles  ABC  and  ADC  are  each  right  angles,  the  sum 
of  the  angles  BAD,  BCD  is  two  right  angles.  But  the 
sum  of  the  adjacent  angles  BCD,  BCD  is  also  two  right 
angles.  Hence,  if  in  these  last  two  sums  we  omit  the  com- 
mon angle  BCD,  we  have  remaining  the  angle  BCD,  equal 
to  the  angle  BAD,  and  consequently  the  angle  BCD  which 
is  the  supplement  of  the  first  of  these  equal  angles  is  also 
the  supplement  of  the  other. 

Hence  the  Theorem. 

ScnoLiUM. — If  the  vertex  of  the  second  angle  be  without  the  first  angle, 
we  would  draw  through  any  assumed  point  within  the  first  angle  parallels  to 
the  sides  of  the  second ;  the  above  demonstration  will  then  apply  to  the  first 
angle,  and  the  angle  formed  by  the  parallels. 

THEOREM    XT.  ,X 

From  any  point  ivithout  a  straight  line,  but  one  perpen- 
dicular can  be  drawn  to  that  line. 

From  the  point  A  let  us  suppose 
it  possible  that  two  perpendiculars, 
A  B  and  A  C,  can  be  drawn.  Now,  be- 
cause AB  is  a  supposed  perpendicu- 
lar, the  angle  ABC  is  a  right  angle  ;  — 
and  because  AC  is  a  supposed  per- 


B   C 


32  GEOMETRY. 

pendicular,  the  angle  ACB  is  also  a  right  angle ;  and  if 
two  angles  of  the  triangle  ABC  are  together  equal  to  two 
right  angles,  the  third  angle,  BAG,  must  be  infinitely 
small,  or  zero  ;  that  is,  the  two  perpendiculars  being  drawn 
through  the  common  point  A,  and  including  no  angle, 
must  necessarily  coincide,  and  form  one  and  the  same  per- 
pendicular. 

Hence  the  theorem  ;  from  any  point  without  a  straight 
line,  etc. 

Cor.  At  a  given  point  in  a  straight  line  but  one  per- 
pendicular can  be  erected  to  that  line ;  for,  if  there  could 
be  two  perpendiculars,  we  should  have  unequal  righ 
angles,  which  is  impossible. 

THEOREM    XVI. 

Two  triangles  which  have  two  sides  and  the  included  angle 
in  the  one,  equal  to  two  sides  and  the  included  angle  in  the 
other,  each  to  each,  are  equal  in  all  respects. 

In  the  two  A's,  ABC  and  BEF, 
on  the  supposition  that  AB  =  BE, 
AC=BF,  and  [_A  =  \__B,  we 
are  to  prove  that  BC  must  =  EF, 
the  [__B  =  [__E,  and  the  [_C  = 

LJJ 

Conceive  the  A  ABC  cut  out  of  the  paper,  taken  up, 
and  placed  on  the  A  BEF  in  such  a  manner  that  the 
point  A  shall  fall  on  the  point  B,  and  the  line  AB  on 
the  line  BE;  then  the  point  B  will  fall  on  the  point  E, 
because  the  lines  are  equal.  Now,  as  the  [__A  =  [__B, 
the  line  A C must  take  the  same  direction  as  BF,  and  fall 
on  BF;  and  as  AC  =  BF,  the  point  C  will  fall  on  F,  B 
being  on  E  and  C  on  F,  BC  must  be  exactly  on  EF, 
(otherwise,  two  straight  lines  would  enclose  a  space,  Ax. 
13),  and  BC=  EF,  and  the  two  magnitudes  exactly  fill 
the  same  space.  Therefore,  BC  =  EF,  [_B  =  [__E, 
_C ~  [_F,  and  the  two  A's  are  equal,  (Ax.  10). 

Hence  the  theorem ;  two  triangles  which  have  two  sides,  evo. 


BOOK    I. 


33 


THEOREM    XVII. 

When  two  triangles  have  a  side  and  two  adjacent  angles  in 
the  one,  equal  to  a  side  and  two  adjacent  angles  in  the  other, 
each  to  each,  the  two  triangles  are  equal  in  all  respects. 

In  two  A's,  as  ABC  and 
T)EF,  on  the  supposition 
th&tBC  =  EF,  \_B=\_E, 
and  [_C  —  [_F,  we  are  to 
prove  that  AB  =  BE,  AC 
=  DF,  andL^.  =  L^>- 

Conceive  the  A  ABC  taken  up  and  placed  on  the  A 
DBF,  so  that  the  side  BC  shall  exactly  coincide  with  its 
equal  side  EF;  now,  because  the  angle  B  is  equal  to  the 
angle  E,  the  line  BA  will  take  the  direction  of  ED,  and 
will  fall  exactly  upon  :t ;  and  because  the  angle  C  is  equal 
to  the  angle  F,  the  line  CA  will  take  the  direction  of 
FD,  and  fall  exactly  upon  it ;  and  the  two  lines  BA  and 
CA,  exactly  coinciding  with  the  two  lines  EB  and  FD, 
the  point  A  will  fall  on  B,  and  the  two  magnitudes  will 
exactly  fill  the  same  space ;  therefore,  by  Ax.  10,  they  are 
equal,  and  AB  »  DE,  AC=DF,  and  the  [_A  =  \__D. 

Hence  the  theorem ;  when  two  triangles  have  a  side  and 
two  adjacent  angles  in  the  one,  equal  to,  etc. 

/\  THEOREM    XVIII. 

If  two  sides  of  a  triangle  are  equal,  the  angles  opposite  to 
these  sides  are  also  equal. 

Let  ABC  be  a  triangle;  and  on 
the  supposition  that  AC  =  BC,  we 
are  to  prove  that  the  l_A=the  [_B. 

Conceive  the  angle  C  divided  into 
two  equal  angles  by  the  line  CD; 
then  we  have  two  A's,  ADC  and 
BDC,  which  have  the  two  sides,  AC 
and  CD  of  the  one,  equal  to  the  two 

sides,  CB  and  CD  of  the  other ;  and 

o 


34  GEOMETRY. 

the  invaded  angle  AOB,  of  the  one,  equa.  to  the  in- 
cluded angle  BOB  of  the  other:  therefore,  (Th.  16),  AB 
=  BB,  and  the  angle  A,  opposite  to  OB  of  the  one  tri- 
angle, is  equal  to  the  angle  B,  opposite  to  OB  of  the 
other  triangle ;  that  is,  |_  A  =  |__  B. 

Hence  the  theorem ;  if  two  sides  of  a  triangle  are  equal, 
the  angles,  etc. 

Cor,  1.  Conversely :  if  two  angles  of  a  triangle  are  equal, 
the  sides  opposite  to  them  are  equal,  and  the  triangle  is 
isosceles. 

For,  if  A 0  is  not  equal  to  BO,  suppose  BO  to  be  the 
greater,  and  make  BE—  AE;  then  will  A  ABB  be  isos- 
celes, and  [_EAB  =  l_EBA  ;  hence  [_EAB  =  [_  CAB, 
or  a  part  is  equal  to  the  whole,  which  is  absurd ;  therefore, 
OB  cannot  be  greater  than  AO,  that  is,  neither  of  the 
sides  AO,  BO,  can  be  greater  than  the  other,  and  conse- 
quently they  are  equal. 

Cor.  2.  As  the  two  triangles,  A  CB  and  BOB,  are  in  all 
respects  equal,  the  line  which  bisects  the  angle  included 
between  the  equal  sides  of  an  isosceles  A  also  bisects  the 
base,  and  is  perpendicular  to  the  base. 

Scholium  1. — If  in  the  perpendicular  DC,  any  other  point  than  C 
be  taken,  and  lines  be  drawn  to  the  extremities  A  and  B,  such  lines 
will  be  equal,  as  is  evident  from  Th.  16 ;  hence,  we  may  announce 
this  truth:  Any  point  in  a  perpendicular  drawn  from  the  middle  of  a 
line,  is  at  equal  distances  from  the  two  extremities  of  the  line. 

ScnoLiuM  2.  —  Since  two  points  determine  the  position  of  a  line,  it 
follows,  that  a  line  which  joins  two  points  each  equally  distant  from  the 
extremities  of  a  given  line,  is  perpendicular  to  this  line  at  its  middle 
point, 

<  K 

r^  THEOREM    XIX. 

The  greater  side  of  every  trianjle  has  the  greater  angle 
opposite  to  it. 

Let  ABC  be  a  A  ;  and  on  the  supposition  that  A  C  is 
greater  than  AB,  we  are  to  prove  that  the  angle  ABCi* 


BOOK    I.  85 

greater  than  the  [_  0.     From  AC,  the 
greater  of  the  two  sides,  take  AB,  equal  ^ 

to  the  less  side  AB,  and  draw  BB,  thus  /\ 

making  two  triangles  of  the  original  tri-  /    \ 

angle.     As  AB  =  AB,  the  [__ABB  =        /       \ 

theL^^A  (Th.18).  B\r~\ 

But  the  |__  ABB  is  the  exterior  angle  ^\\ 

of  the  A  BBC,  and  is  therefore  greater  c 

than    C,  (Th.   12,  Cor.   1);    that  is,  the 
L  ABD  is  greater  than  the  [_  C.     Much  more,  then,  is 
the  angle  ABC  greater  than  the  angle  C. 

Hence  the  theorem ;  the  greater  side  of  every  triangle,  etc. 

Cor,  Conversely:  the  greater  angle  of  any  triangle  has 
the  greater  side  opposite  to  it. 

In  the  triangle  ABC,  let  the  angle  B  be  greater  than 
the  angle  A ;  then  is  the  side  AC  greater  than  the  side 
BC 

For,  if  BC  —  A C,  the  angle  A  must  be  equal  to  the 
angle  B,  (Th.  18),  which  is  contrary  to  the  hypothesis ; 
and  if  BC^>AC,  the  angle  A  must  be  greater  than  the 
angle  B,  by  what  is  above  proved,  which  is  also  contrary 
to  the  hypothesis ;  hence  BC  can  be  neither  equal  to,  nor 
greater,  than  AC;  it  is  therefore  less  than  AC. 

THEOREM    XX. 

The  difference  between  any  two  sides  of  a  triangle  is  less 
than  thi  third  side. 

Let  ABC  he  a  A,  in  which  A  C  is  greater 
than  AB ;  then  we  are  to  prove  that  A  C 
—AB  is  less  than  BC. 

On  AC,  the  greater  of  the  two  sides, 
lay  off  AB  equal  to  AB. 

Now,  as  a  straight  line  is  the  shortest 
distance  between  two  points,  we  have 

AB  +  BOAC.  (1) 


36 


GEOMETRY. 


x 


From  these  unequals  suotract  the  equals  AB  =  AB, 
and  we  have  BO>AO—  AB.     (Ax.  5). 

Hence  the  theorem ;  the  difference  between  any  two  Met 
of  a  triangle,  etc.  -^ 

\ 
THEOREM    XXI. 


ff  two  triangles  have  the  three  sides  of  the  one  equal  to 
the  three  sides  of  the  other,  each  to  each,  the  two  triangles  are 
equxl,  and  the  equal  angles  are  opposite  the  equal  sides. 

In  two  triangles,  as  ABO  and  ABB,  on  the  supposition 
that  the  side  AB  of  the  one  =  the  side  AB  of  the  other, 
AO=AB,  and  BO=BJ),  we  are  to  demonstrate  that 
[_AOB  =  \__ABB,  L  BAO  = 
l_BAB,  and  [_ABO=  [_ABD. 

Conceive  the  two  triangles  to 
t>e  joined  together  by  their  long- 
est equal  sides,  and  draw  the 
line  OB. 

Then,  in  the  triangle  AOD, 
because  AO  is    equal   to  AB, 

the  angle  AOD  is  equal  to  the  angle  ABO,  (Th.  18).  In 
like  manner,  in  the  triangle  BOB,  because  BO  is  equal 
to  BB,  the  angle  BOB  is  equal  to  the  angle  BBO.  Now, 
the  angle  AOB  being  equal  to  the  angle  ABO,  and  the 
angle  BOB  to  the  angle  BBO,  [_AOB  +  [_BOB  =  L. 
ABO  +  [_BBO,  (Ax.  2) ;  that  is,  the  whole  angled  OB  ia 
equal  to  the  whole  angle  ABB. 

Since  the  two  sides  A  0  and  OB  are  equal  to  the  two  sides 
AB  and  BB,  each  to  each,  and  their  included  angles  A  OB, 
ABB,  are  also  equal,  the  two  triangles  ABO,  ABB,  are 
equal,  (Th.  16),  and  have  their  other  angles  equal ;  that 
is,  [_BAO=  \_BAB,  and  [_ABO=  [__ABB. 

Hence  the  theorem ;  if  two  triangles  have  the  three  siie* 
of  the  one,  etc. 


BOOK    I 


87 


THEOREM    XXII. 

If  two  triangles  have  two  sides  of  the  one  equal  to  tw* 
tides  of  the  other,  each  to  each,  and  the  included  angles  ui> 
equal,  the  third  sides  will  be  unequal,  and  the  greater  third 
side  will  belong  to  the  triangle  which  has  the  greater  included 
angle. 

In  the  two  A's,  ABO  and 
ACB,  let  AB  and  AC  of  the 
one  A  be  equal  to  AB  and  A  0 
of  the  other  A,  and  the  angle 
BAC  greater  than  the  angle 
BAG;  we  are  to  prove  that 
the  side  BO  is  greater  than  the 
side  OB. 

Conceive  the  two  a's  joined  together  by  their  shorter 
equal  sides,  and  draw  the  line  BB.  Now,  as  AB  =  AB, 
ABB  is  an  isosceles  A.  From  the  vertex  A,  draw  a  line 
bisecting  the  angle  BAB.  This  line  must  be  perpendic- 
ular to  the  base  BB,  (Th.  18,  Cor.  2).  Since  the  \_BAO 
is  greater  than  the  \_BAO,  this  line  must  meet  BO,  and 
will  not  meet  OB.  From  the  point  E,  where  the  per- 
pendicular meets  BO,  draw  BB. 

Now  BB  -  BB,  (Th.  18,  Scholium  1). 

Add  BO  to  each ;  then  BO  =  BE  +  EC. 

But  BE  +  EO  is  greater  than  BO. 

Therefore  BOy  BO. 

Hence  the  theorem ;  if  two  triangles  have  two  sides  of 
one  equal  to  two  sides  of  the  other,  etc. 

Cor.  Any  point  out  of  the  perpendicular  drawn  from 
the  middle  point  of  a  line,  is  unequally  distant  from  the 
extremities  of  the  line. 


-f 


y 


GEOMETRY. 


THEOREM    XXIII, 


A  perpendicular  is  the  shortest  line  that  can  be  drawn  from 
any  point  to  a  straight  line  ;  and  if  other  lines  be  drawn  from 
the  same  point  to  the  same  straight  line,  that  which  meets  it 
farthest  from  the  perpendicular  will  be  longest ;  and  lines 
at  equal  distances  from  the  perpendicular,  on  opposite 
sides,  are  equal. 

Let  A  be  any  point  without  the 
line  BE ;  let  AB  be  the  perpen- 
dicular; and  AC,  AB,  and  AE 
oblique  lines:  then,  if  BO  is  less 
than  BB,  and  BO—  BE,  we  are  to 
show, 

1st.  That  AB  is  less  than  A  C 
2d.  That  A 0  is  less  than  AB.     3d.  That  AC=  AE. 

1st.  In  the  triangle  ABO,  as  AB  is  perpendicular  to 
BC,  the  angle  ABC  is  a  right  angle;  and,  therefore  (by 
Theorem  12,  Cor.  4) ;  the  angle  BCA  is  less  than  a  right 
angle ;  and,  as  the  greater  side  is  always  opposite  the 
greater  angle,  AB  is  less  than  AC;  and  AC  may  be 
any  line  not  identical  with  AB ;  therefore  a  perpen- 
dicular is  the  shortest  line  that  can  be  drawn  from  A 
to  the  line  DE. 

2d.  As  the  two  angles,  A  OB  and  A  CD,  are  together 
equal  to  two  right  angles,  (Th.  1),  and  A  OB  is  less  than 
a  right  angle,  AOB  must  be  greater  than  a  right  angle ; 
consequently,  the  [__  B  is  less  than  a  right  angle ;  and,  in 
the  A  AOB,  AB  is  greater  than  AC,  or  A 0  is  less  than 
AB,  (Th.  19  Cor). 

3d.  In  the  A's  ABC  and  ABE,  AB  is  common,  CB=> 
BE,  and  the  angles  at B are  right  angles;  therefore,  AC  — 
AE,  (Th.  16). 

Hence  the  theorem;  a  perpendicular  is  the  shortest  line 
etc. 

Cor.  Conversely :  if  two  equal  oblique  lines  be  drawn 


BOOK    I.  39 

from  the  same  point  to  a  given  straight  line,  they  will 
meet  the  line  at  equal  distances  from  the  foot  of  the  per- 
pendicular drawn  from  that  point  to  the  given  line. 

THEOREM   XXIV. 

The  opposite  sides,  and  also  the  opposite  angles  of  any  par- 
allelogram, are  equal. 

Let  ABOD  be  a  parallelogram. 
Then  we  are  to  show  that  AB  =  1)0, 
AD  =  BO,  [__A  -  L  °>  and  [_ADC 
«  \__ABO. 

Draw  a  diagonal,  as  BD;  now,  be- 
cause AB  and  DC  are  parallel,  the  al- 
ternate angles  ABB  and  BBC  are  equal,  (Th.  6).     Foi 
the  same  reason,  as  AB  and  BO  are  parallel,  the  angles 
ABB  and  BBO  are  equal.     !Now,  in  the  two  triangles 
ABB  and  BOB,  the  side  BD  is  common, 

the  \_ABB  =  [_DBO  (1) 

tmd[_BDO  =  \_ABD  (2) 

Therefore,  the  angle  A  =  the  angle  C,  and  the  two  tri- 
angles are  equal  in  all  respects,  (Th.  17);  that  is,  the 
sides  opposite  the  equal  angles  are  equal ;  or,  AB  —  DO, 
and  AD  =  BO.  By  adding  equations  ( 1 )  and  ( 2 )?  we  have 
the  angle  ADO=  the  angle  ABO,  (Ax.  2). 

Hence  the  theorem ;  the  opposite  sides,  and  the  opposite 
angles,  etc. 

Cor.  1.  As  the  sum  of  all  the  angles  of  a  parallelogram 
is  equal  to  four  right  angles,  and  the  angle  A  is  always 
equal  to  the  opposite  angle  O;  therefore,  if  A  is  a  right 
angle,  0  is  also  a  right  angle,  and  the  figure  is  a  rect- 
angle. 

Oor.  2.  As  the  angle  ABO,  added  to  the  angle  A,  gives 
the  same  sum  as  the  angles  of  the  /\ADB;  therefore, 
the  two  adjacent  angles  of  a  parallelogram  are  together 
equal  to  two  right  angles. 


-f 


X 


40  GEOMETRY. 

THEOREM   XXV. 

If  the  opposite  sides  of  a  quadrilateral  are  equal,  they  are 
also  parallel,  and  the  figure  is  a  parallelogram. 

Let  A  BCD  be  any  quadrilateral; 
on  the  supposition  that  AD  =  BC,  and 
AB  =*  DC.  we  are  to  prove  that  AD  is 
parallel  to  BC,  and  AB  parallel  to  DC 

Draw  the  diagonal  BD;  we  now 
have  two  triangles,  ABD  and  BOB, 
which  have  the  side  BD  common,  AD  of  the  one  =.  BO 
of  the  other,  and  AB  of  the  one  =  CD  of  the  other ; 
therefore  the  two  A's  are  equal,  (Th.  21),  and  the 
angles  opposite  the  equal  sides  are  equal ;  that  is,  the 
angle  ADB  =  the  angle  CBD ;  but  these  are  alternate 
angles;  hence,  AD  is  parallel  to  BC,  (Th.  7,  Cor.  1); 
and  because  the  angle  ABD  =  the  angle  BDC,  AB  is 
parallel  to  CD,  and  the  figure  is  a  parallelogram. 

Hence  the  theorem ;  if  the  opposite  sides  of  a  quadri- 
lateral, etc. 

Cor.  This  theorem,  and  also  Th.  24,  proves  that  the 
two  A's  which  make  up  the  parallelogram  are  equal; 
and  the  same  would  be  true  if  we  drew  the  diagonal 
from  A  to  C;  therefore,  the  diagonal  of  any  parallelogram 
bisects  the  parallelogram. 

THEOREM   XXVI. 

The  lines  which  join  the  corresponding  extremities  of  two 
equal  and  parallel  straight  lines,  are  themselves  equal  and 
parallel;  and  the  figure  thus  formed  is  a  parallelogram. 

On  the  supposition  that  AB  is 
equal  and  parallel  to  DC,  we  are  to 
prove  that  AD  is  equal  and  parallel 
to  BC;  and  that  the  figure  is  a  par- 
allelogram. 

Draw  the  diagonal  BD  ;  now,  since 


BOOK   1.  41 

AB  and  DO  are  parallel,  and  BB  joins  them,  the  alter- 
nate angles  ABB  and  BBO  are  equal ;  and  since  the 
side  AB  =  the  side  BO,  and  the  side  BB  i«  common  to 
the  two  A's  ABB  and  OBB,  therefore  the  two  triangles 
are  eqnal,  (Th.  16) ;  that  is,  AB  —BO,  the  angle  A  =  0, 
and  the  \__ABB  =  the  [_BBO;  also  AB  is  parallel  to 
BO;  and  the  figure  is  a  parallelogram. 

Hence  the  theorem ;  the  lines  which  join  the  corresponding 
extremities,  etc. 

THEOREM   XXVII. 

Parallelograms  on  the  same  base,  and  between  the  same 
parallels,  are  equivalent,  or  equal  in  respect  to  area  or  sur- 
face. 

Let  ABEO  and  ABBF  be  two 
parallelograms  on  the  same  base 
AB,  and  between  the  same  paral- 
lels AB  and  OB  ;  we  are  to  prove 
that  these  two  parallelograms  are 
equal. 

Now,  OB  and  FB  are  equal,  be- 
cause they  are  each  equal  to  AB,  (Th.  24) ;  and,  if  from 
the  whole  line  OB  we  take,  in  succession,  OB  and  FB, 
there  will  remain  EB  =  OF,  (Ax.  3) ;  but  BB  =  A 0,  and 
AF=  BB,  (Th.  24);  hence  we  have  two  A's,  OAF  and 
EBB,  which  have  the  three  sides  of  the  one  equal  to  the 
three  sides  of  the  other,  each  to  each ;  therefore,  the  tvro 
A's  are  equal,  (Th.  21).  If,  from  the  whole  figure 
A  BBO,  we  take  away  the  A  OAF,  the  parallelogram 
ABBF  will  remain  ;  and  if  from  the  whole  figure  we  take 
away  the  other  A  EBB,  the  parallelogram  ABEO  will 
remain.  Therefore,  (Ax.  3),  the  parallelogram  ABBF  = 
the  parallelogram  ABEO. 

Hence  the  theorem ;  Parallelograms  on  the  same  base,  etc. 
4* 


42  GEOMETRY. 

THEOREM    XXVIII. 

Triangles  on  the  same  base  and  between  the  same  parallel* 
are  equivalent. 

Let  the  two  A's  ABE  and  ABF 
have  the  same  base  AB,  and  be  be- 
tween the  same  parallels  AB  and 
EF',  then  we  are  to  prove  that  they 
are  equal  in  surface. 

From  B  draw  the  line  BB,  par- 
allel to  AF;  and  from  A  draw  the  line  AC,  parallel  to 
BE ;  and  produce  EF,  if  necessary,  to  C  and  D ;  now  the 
parallelogram  ABBF  =  the  parallelogram  ABEC,  (Th. 
27).  But  the  A  ABE  is  one  half  the  parallelogram 
ABEC,  and  the  A  ABF  is  one  half  the  parallelogram 
ABDF;  and  halves  of  equals  are  equal,  (Ax.  T);  there-, 
fore  the  A  ABE  -  the  A  ABF. 

Hence  the  theorem ;  triangles  on  the  same  base,  etc. 

THEOREM    XXIX. 

Parallelograms  on  equal  bases,  and  between  the  same  par 
allels,  are  equal  in  area. 

Let  ABCB  and  EFaH,  be  two 
parallelograms  on  equal  bases,  AB 
and  EF,  and  between  the  same 
parallels,  AF  and  BG ;  then  Ve  are 
to  prove  that  they  are  equal  in  area. 

AB  =  EF=HG;  but  lines  which  join  equal  and 
parallel  lines,  are  themselves  equal  and  parallel,  (Th.  26) ; 
therefore,  if  AH  said  BGr  be  drawn,  the  figure  ABGrlTis 
a  parallelogram  ==  to  the  parallelogram  ABCB,  (Th.  27) ; 
and  if  we  turn  the  whole  figure  over,  the  two  parallelo- 
grams, G-HEF  and  GRAB,  will  stand  on  the  same  base, 
GH,  and  between  the  same  parallels ;  therefore,  GHEF 
-  GHAB,  and  consequently  ABCB  =  EFGff,  (Ax.  1). 

Hence  the  theorem ;  Parallelograms  on  equal  bases,  (tc. 


DC  H        G 


BOOK   I.  I —       43 

Cor.  Triangles  on  equal  bases,  and  between  the  same 
parallels,  are  equal  in  area.  For,  draw  BB  and  E Gr\  the 
A  ABB  is  one  half  of  the  parallelogram  AC,  and  the 
A  EFGr  is  one  half  of  the  equivalent  parallelogram  FE; 
therefore,  the  A  ABB  -  the  A  EFG,  (Ax.  7). 

THEOREM   XXX. 

If  a  triangle  and  a  parallelogram  are  upon  the  same  or  equal 
bases,  and  between  the  same  parallels,  the  triangle  is  equiva- 
lent to  one  half  the  parallelogram. 

Let  ABC  be  a  A,  and  ABBE  a 
parallelogram,  on  the  same  base  AB, 
and  between  the  same  parallels ;  then 
we  are  to  prove  that  the  A  ABC  is 
equivalent  to  one  half  of  the  parallel- 
ogram ABBE. 

Draw  EB  the  diagonal  of  the  parallelogram ;  now, 
because  the  two  A's  ABC  and  ABE  are  on  the  same 
base,  and  between  the  same  parallels,  they  are  equiva- 
lent, (Th.  28) ;  but  the  A  ABE  is  one  half  the  parallel- 
ogram ABBE,  (Th.  25,  Cor.) ;  therefore  the  A  ABC  is 
equivalent  to  one  half  of  the  same  parallelogram,  (Ax.  7). 

Hence  the  theorem ;  if  a  triangle  and  a  parallelogram, 
etc 

THEOREM   XXXI. 

The  complementary  parallelograms  described  about  any 
point  in  the  diagonal  of  any  parallelogram,  are  equivalent  to 
each  other. 

Let  A  C  be  a  parallelogram,  and 
BB  its  diagonal ;  take  any  point, 
as  E,  in  the  diagonal,  and  through 
this  point  draw  lines  parallel  to  the 
sides  of  the  parallelogram,  thus 
forming  four  parallelograms. 

We  are  now  to  prove  that  the  complementary  paral- 
lelograms, AE  and  EC,  are  equivalent. 


44  GEOMETRY. 

By  (Th.  25,  Cor.)  we  learn  that  the  A  ABB  =  A  BBC. 
Also  by  the  same  Cor.,  A  a  =  A  b,  and  Ac  =  A<i;  there- 
fore by  addition 

Aa-fAc  =  A5  +  Ac?. 

Now,  from  the  whole  A  ABB  take  A  a  +  A  e,  and 
from  the  whole  A  BBC  take  the  equal  sum,  A  b  +  A  d, 
and  the  remaining  parallelograms  AE  and  EC  are  equiv~ 
alent,  (Ax.  3). 

Hence  the  theorem ;  the  complementary  parallelogram^ 
etc. 

THEOREM  XXXII. 

The  perimeter  of  a  rectangle  is  less  than  that  of  any  rhom- 
hoid  standing  on  the  same  base,  and  included  between  the  same 
parallels. 

Let  ABCB  be  a  rect- 
angle, and  ABEF  a  rhom- 
boid having  the  same  base, 
and  their  opposite  sides 
in  the  same  line  parallel 
to  the  base. 

We  are  now  to  prove  that  the  perimeter  ABCB  A  is  less 
than  ABEFA. 

Because  AB  is  a  perpendicular  from  A  to  the  line  BEt 
and  AF  an  oblique  line,  AB  is  less  than  AF,  (Th.  23). 
For  the  same  reason  BC is  less ^than  BE;  hence  AT) + 
BC<AF+  BE.  Adding  the  sum,  AB  +  BC,  to  the  first 
member  of  this  inequality,  and  its  equal  AB  4-  FE  to 
the  second  member,  we  have  AB  +  BC  +  CD  +  BA,  or 
the  perimeter  of  the  rectangle,  less  than  AB  -f  BE  -f 
EF  -f  FA,  or  the  perimeter  of  the  rhomboid.  Hen  ?e 
the  theorem;  the  perimeter  of  a  rectangle,  etc. 


BOOK  I 


45 


Thus  far,  areas  have  "been  considered  only  relatively 
and  in  the  abstract.  We  will  now  explain  how  we  may 
pass  to  the  absolute  measures,  or,  more  properly,  to  the 
numerical  expressions  for  areas. 

THEOREM  XXXIII. 

The  area  of  any  plane  triangle  is  measured  by  the  pro* 
duct  of  its  base  by  one  half  its  altitude  ;  or  by  one  half  of 
the  product  of  its  base  by  its  altitude. 

Let  ABO  represent  any  triangle,  AB 
its  base,  and  AD,  at  right  angles  to  AB, 
its  altitude ;  now  we  are  to  show  that  the 
area  of  A  BO  is  equal  to  the  product  of 
AB  by  one  half  of  AB ;  or  one  half  of 
AB  by  AB ;  or  one  half  of  the  product  of  AB  by  AB. 

On  AB  construct  the  rectangle  ABED;  and  the  area 
of  this  rectangle  is  measured  by  AB  into  AB  (Def. 
54) ;  but  the  area  of  the  A  ABO  is  equivalent  to  one 
half  this  rectangle,  (Th.  30).  Therefore,  the  area  of  the 
A  is  measured  by  }  AB  x  AB,  or  one  half  the  product 
of  its  base  by  its  altitude.  Hence  the  theorem ;  the  area 
of  any  plane  triangle,  etc. 

THEOREM  XXXIV. 

The  area  of  a  trapezoid  is  measured  by  one  half  the  mm 
of  its  parallel  sides  multiplied  by  the  perpendicular  distance 
between  them. 

Let  ABBQ  represent  any  trape- 
zoid; draw  the  diagonal  BO,  divid- 
ing it  into  two  triangles,  ABO 'and 
BOB:  OB  is  the  base  of  one  tri- 
angle, and  AB  may  be  considered 
as  the  base  of  the  other ;  and  EF  is  the  common  altitude 
of  the  two  triangles. 

Now,  by  Th.  33,  the  area  of  the  triangle  BOB  =-\OB 
v  EF;  and  the  area  of  the  A  ABO=  \AB  X  EF;  but 


46 


GEOMETRY. 


L     K       I 


II 


by  addition,  the  area  of  the  two  A's,  or  of  the  trape* 
zoid,  is  equal  to  i  (AB+  CD)xEF.  Hence  the  theorem; 
the  area  of  a  trapezoid,  etc. 

THEOREM  XXXV. 

If  one  of  two  lines  is  divided  into  any  number  of  parts,  the 
rectangle  contained  by  the  two  lines  is  equal  to  the  sum  of  the 
several  rectangles  contained  by  the  undivided  line  and  the  seve- 
ral parts  of  the  divided  line. 

Let  AB  and  AD  be  two  lines, 
and  suppose  AB  divided  into  any 
number  of  parts  at  the  points  E, 
F,  Gr,  etc. ;  then  the  whole  rect- 
angle contained  by  the  two  lines 
is  AH,  which  is  measured  by  AB 
into  AD.  But  the  rectangle  AL  is  measured  by  AB 
into  AD ;  the  rectangle  EK  is  measured  by  EF  into  EL, 
which  is  equal  to  EF  into  AD;  and  so  of  all  the  other 
partial  rectangles ;  and  the  truth  of  the  proposition  is  as 
obvious  as  that  a  whole  is  equal  to  the  sum  of  all  its 
parts.  Hence  the  theorem ;  if  one  of  two  lines  is  divided, 
etc.  ^/f  ^  Uc  , 

v.  One*.    0 

X  THEOREM    XXXVI. 


G       B 


X 


C      B 


If  a  straight  line  is  divided  into  any  two  parts,  the  square 
described  on  the  whole  line  is  equivalent  to  the  sum  of  the 
squares  described  on  the  two  parts  plus  twice  the  rectangle  con- 
tained by  the  parts. 

Let  AB  be  any  line  divided  into 
any  two  parts  at  the  point  0;  now  we 
are  to  prove  that  the  square  on  AB 
is  equivalent  to  the  sum  of  the 
squares  on  A  0  and  OB  plus  twice  the 
rectangle  contained  by  AC  and  CB. 

On  AB  describe  the  square  AD. 
Through  the  point  0  draw  QM,  par- 


M    D 


BOOK    I.  47 

allel  to  BD ;  take  BE  =  BO,  and  through  E  draw  EKN, 
parallel  to  AB.  We  now  have  OH,  the  square  on  OB,  by 
direct  construction. 

As  AB  =  BD,  and  OB  -  £.0",  by  subtraction,  AB  — 
OB  =  BD  —  BE;  or  A0=  ED.  But  NK=AO,  being 
opposite  sides  of  a  parallelogram ;  and  for  the  same  rea- 
son, KM  =  I7D.  Therefore,  (Ax.  1),  NK  =  KM,  and  the 
figure  NM is  a  square  on  NK,  equal  to  a  square  on  .AG7. 
But  the  whole  square  on  AB  is  composed  of  the  two 
squares  OE,  NM,  and  the  two  complements  or  rectangles 
AK  and  KB  ;  and  since  each  of  these  latter  is  AC  in 
length,  and  BO  in  width,  each  has  for  its  measure  AC  into 
(7i?;  therefore  the  whole  square  on  AB  is  equivalent  to 
AC2  +  BC2  +  2A0x  OB. 

Hence  the  theorem ;  if  a  straight  line  is  divided  into  any 
two  parts,  etc. 

This  theorem  may  be  proved  algebraically,  thus : 

Let  w  represent  any  whole  right  line  divided  into  any 
two  parts  a  and  b ;  then  we  shall  have  the  equation 

w  =  a  +  b 
By  squaring,  w2  =  a2  -f  b2  -f  2ab. 

Oor.  If  a  =  b,  then  w2  =  4a2 ;  that  is,  the  square  de- 
scribed on  any  line  is  four  times  the  square  described  on 
one  half  of  it. 

THEOREM    XXXVII. 

The  square  described  on  the  difference  of  two  lines  is  equiv- 
alent to  the  sum  of  the  squares  described  on  the  two  lines  di- 
minished by  twice  the  rectangle  contained  by  the  lines. 

Let  AB  represent  the  greater  of  two  lines,  OB  the 
Jess  line,  and  A  0  their  difference. 

We  are  now  to  prove  that  the  square  described  on  A  C 
is  equivalent  to  the  sum  of  the  squares  on  AB  and  BC 
diminished  by  twice  the  rectangle  contained  by  AB 
and  BO. 

Conceive  the  square  AF  to  be  described  on  AB,  and 


L     E 


48  GEOMETRY. 

the  square  BL  on  CB ;  on  AC  describe        H  F 

the  square  ACGM,  and  produce  MG 

to  K.  M 

As  GC=AC,  and  CL  ^  CB,  by 
addition,  (GC  +  CL),  or  GL,  is  equal 
to  AC  +  CB,  or  J.jB.  Therefore,  the 
rectangle  GE  is  AB  in  length,  and 
Oi?  in  width,  and  is  measured  by  AB 

y.BC. 

Also  AH=  AB,  and  AM  =  J.  (7;  by  subtraction,  Jfff 

=  <7i?;  and  as  MK  =  AB,  the  rectangle  HK  is  J.i?  in 
length,  and  CB  in  width,  and  is  measured  by  AB  x  BC; 
and  the  two  rectangles  GE  and  .ffif  are  together  equiva- 
lent to  2AB  x  BC. 

Now,  the  squares  on  AB  and  BC  make  the  whole  figure 
AHFELC;   and  from  this  whole  figure,  or  these  two 
squares,  take  away  the  two  rectangles  HK  and  G  E,  and 
the  square  on  A C only  will  remain;  that  is, 
AC2  =  AB*  +  B~C*  —  2AB  x  BC 

Hence  the  theorem ;  the  square  described  on  the  differ 
ence  of  two  lines,  etc. 

This  theorem  may  be  proved  algebraically,  thus : 

Let  a  represent  the  greater  of  two  lines,  b  the  less,  and 
d  their  difference ;  then  we  must  have  this  equation : 
d  =  a  — b 
By  squaring,    d2  =  a2  +  b2  —  2ab. 

a  a2 

Cor.  If  d  =  b,  then  d  =  -~,  and  d2  =  -r ;    that  is,   the 

equare  described  on  one  half  of  any  line  is  equivalent 
to  one  fourth  of  the  square  described  on  the  whole  line. 

THEOREM   XXXVIII. 

The  difference  of  the  squares  described  on  any  two  lines  is 
equivalent  to  the  rectangle  contained  by  the  sum  and  difference 
of  the  lines. 

Let  A  B  be  the  greater  of  two  lines,  and  A  C  the  less, 
and  on  these  lines  describe  the  squares  AB,  AM;  then,  the 


BOOK   I.  49 

difference  of  the  squares  on  AB  and  A  0  is  the  two  rect- 
angles EF  and  FO.     We  are  now  to 
show  that  the  measure  of  these  reck 
angles  may  be  expressed  by  (AB  +  AC) 
x(AB—AO). 

i     The  length  of  the  rectangle  EF  is  ED, 
or  its  equal  AB;  and  the  length  of  the 
rectangle  FO  is  MO,  or  its  equal  AO; 
therefore,  the  length  of  the  two  together  (if  we  con 
ceive  them  put  between  the  same  parallel  lines)  will  be 
AB-\-  AO;  and  the  common  width  is  OB,  which  is  equal 
to  AB—AO;  therefore,  AB2—  A02  =  (AB+AO)  x  (AB 
—AO). 

Hence  the  theorem;  the  difference  of  the  squares  de- 
scribed on  any  two  lines,  etc. 

This  theorem  may  be  proved  algebraically:  thus, 

Let  a  represent  one  line,  and  b  another ; 

Then  a  -f  b  is  their  sum,  and  a  —  b  their  difference ; 

and  (a  +  b)  X  (a  —  b)  =  a2  —  b\ 

THEOREM   XXXIX. 

The  square  described  on  the  hypotenuse  of  any  right-angled 
triangle  is  equivalent  to  the  sum  of  the  squares  described  on 
the  other  two  sides. 

Let  ABO  represent  any  righkangled  triangle,  the  right 
angle  at  B;  we  are  to  prove  that  the  square  on  A  0  is 
equivalent  to  the  sum  of  two  squares;  one  on  AB,  the 
other  on  BO. 

On  the  three  sides  of  the  triangle  describe  the  three 
squares,  AB,  AL  and  BM.  Through  the  point  B,  draw 
BNE  perpendicular  to  AO,  and  produce  it  to  meet  tho 
line  Grim  K;  also  produce  AF  to  meet  CrI  in  S,  and 
ML  to  meet  GI  produced  in  K. 

Remark.  —  That  the  lines,  GI  and  ML,  produced,  meet  at  the  point 
BT,  may  be  readily  shown.     As  the  proof  of  this  fact  is  not  necessary  for 
tht-  demonstration,  it  is  left  as  an  exercise  for  the  learner. 
5  D 


50 


GEOMETRY. 


The  angle  BAG  is  a  right  angle,  and  the  angle  NAH 
is  also  a  right  angle ;  if 
from  these  equals  we 
subtract  the  common 
angle  BAH,  the  re- 
maining angle,  BAO, 
must  be  equal  to  the  re- 
maining angle  &AH. 
The  angle  G-  is  a  right 
angle,  equal  to  the 
angle  ABC;  and  AB 
=  AGr;  therefore,  the 
two  A's  ABO  and 
AGrH  are  equal,  and 
AH=AQ.  But^<7  = 
AF;  therefore,  AH= 
AF.      Now,   the    two 

parallelograms,  AE  and  AHKB  are  equivalent,  because 
they  are  upon  equal  bases,  and  between  the  same  paral- 
lels, Zffand  UK,  (Th.  29). 

But  the  square  AI,  and  the  parallelogram  AHKB,  are 
equivalent,  because  they  are  on  the  same  base,  AB,  and 
between  the  same  parallels,  AB  and  GrK;  therefore,  the 
square  AI,  and  the  parallelogram  AF,  being  each  equiv- 
alent to  the  same  parallelogram  AHKB,  are  equivalent 
to  each  other,  (Ax.  1).  In  the  same  manner  we  may 
prove  that  the  square  BM  is  equivalent  to  the  rectangle 
ND ;  therefore,  by  addition,  the  two  squares,  AI  and 
BM,  are  equivalent  to  the  two  parallelograms,  AF  and 
ND,  or  to  the  square  AD. 

Hence  the  theorem ;  the  square  described  on  the  hypoU 
nuse  of  a  right-angled  triangle,  etc. 

Cor.  If  two  right-angled  triangles  have  the  hypotenuse,  and 
a  side  of  the  one  equal  to  the  hypotenuse  and  a  side  of  the 
other,  each  to  each,  the  two  triangles  are  equal. 


BOOK  I.  51 

Let  ABO  ard  AGS  be  the  two  A's,  in  which  we  sap- 
pose  AC  =  AH,  and  BC  =  GH;  then  will  AG  =  AB 

For,  we  have  AC2  =  AB2  +  ff(72, 

or,  by  transposing,  AC2  —  BO2  =  ~A&, 

and  AE2  =  ~AG2+G1I2, 

or,  by  transposing,  A H2 —  GH2  =  AG2. 

But  by  the  hypothesis  AC2  —~BQ2  =  AS*  —  (JS* ; 

hence,     AB2  =  J.  G2,  or,  ^ J5  =  A  G. 

Scholium. — The  two  sides,  AB  and  BC,  may  vary,  while  A  C  remain* 
constant.  AB  may  be  equal  to  BC\  then  the  point  JVwill  be  in  th* 
middle  of  A  C.  When  AB  is  very  near  the  length  of  A  C,  and  BC  very 
email,  then  the  point  N  falls  very  near  to  C.  Now  as  AE  and  ND  are 
right-angled  parallelograms,  their  areas  are  measured  by  the  product 
of  their  bases  by  their  altitudes ;  and  it  is  evident  that,  as  they  have  the 
same  altitude,  these  areas  will  vary  directly  as  their  bases  AN  and 
NC;  hence  the  squares  on  AB  and  BC,  which  are  equivalent  to  those 
rectangles,  vary  as  the  lines  AN  and  NC. 

The  following  outline  of  the  demonstration  of  this  pro- 
position is  presented  as  a  useful  disciplinary  exercise  for 
the  student. 

We  employ  the  same  figure,  in  which  no  change  is 
made  except  to  draw  through  0  the  line  OP,  parallel  to  BK. 

The  first  step  is  to  prove  the  equality  of  the  triangles 
AGE  and  ABO,  whence  AH  =  AC.  But  AO  =  AF; 
therefore  AH=  AF. 

The  parallelograms  AFEJSF  and  AEKB  are  equiva 

lent.  Also,  the  parallelogram  AHKB  =  the  square  ABIG, 

(Th.  27),  and  the  parallelogram  KBCP=  NEDC=  square 

BOML.     Now,  by  adding  the  equals 

AFEN=  ABIG 

JSTEBO  =  BOML 

we  obtain  AFDO  =  ABIG  +  BOML. 

That  is,  the  square  on  AO  is  equivalent  to  the  sum  of 
the  squares  on  AB  and  BO. 

The  great  practical  importance  of  this  theorem,  in  the 
extent  and  variety  of  its  applications,  and  the  frequency 
of  its  use  in  establishing  subsequent  propositions,  ren- 
ders it  necessary  that  the  student  should  master  it  com- 
pletely.    To  secure  this  end,  we  present  a 


52 


GEOMETRY. 


Second  Demonstration 

Let  ABO  be  a  triangle 
right-angled  at  B.  On  the 
hypotenuse  AO,  describe  the 
square  A  OEB.  From  B  and 
E  let  fall  the  perpendiculars 
Bb  and  Ed,  on  J.1?  and  AB 
produced.  Draw  Bn  and  Ca, 
making  right  angles  with 
Ed. 

We  give  an  outline  only 
of  the  demonstration,  requiring  the  pupil  to  make  it 
complete. 

First  Part. — Prove  the  four  triangles  ABO,  AbB,  BnE, 
and  EaO,  equal  to  each  other. 

The  proof  is  as  follows:  The  A's  ABO  and  BnE  are 
equal,  because  the  angles  of  the  one  are  equal  to  the 
angles  of  the  other,  each  to  each,  and  the  hypotenuse 
AO  of  the  one,  is  equal  to  the  hypotenuse  BE  of  the 
other.  In  like  manner,  it  may  be  shown  that  the  a's 
AbB  and  EaO  are  equal. 

Now,  the  sum  of  the  three  angles  about  A,  is  equal  to 
the  sum  of  the  three  angles  of  the  A  ABO;  and  if,  from 
the  first  sum,  we  take  \__BAO  +  [__  OAB,  and  from  the 
second  we  take  L^  +  L  GAB  ==  [_BAO+  [_  OAB,  the 
remaining  angles  are  equal ;  that  is,  [__  BAb  is  equal  to 
\__AOB ;  hence  the  A's  ABO  and  BbA  have  their  angles 
equal,  each  to  each;  and  since  A (7=  BA,  the  A's  are 
themselves  equal,  and  the  four  triangles  ABO,  AbB, 
DnE,  and  EaO,  are  equal  to  each  other. 

Second.  —  Prove  that  the  square  bBnd  is  equal  to  a 
iquare  on  AB.     The  square  BdaO  is  obviously  on  BO. 

Third. — The  area  of  the  whole  figure  is  equal  to  the 
square  on  AO,  and  the  area  of  two  of  the  four  equal 
right-angled  triangles. 

Also,  the  area  of  the  whole  figure  is  equal  to  two  other 


BOOK    1.  53 

Bquares,  bDnd  and  daOB,  and  two  of  the  tour  equal  tri- 
angles, DnE  and  EaO, 

Omitting  or  subtracting  the  areas  of  two  of  the  four 
right-angled  a's  from  each  of  the  two  expressions  for 
the  area  of  the  whole  figure,  there  will  remain  the  square 
on  AO,  equal  to  the  sum  of  the  two  squares,  Dndb  and 
daCB. 

That  is,  AB*  +  BO*  =  AC*. 

Hence  the  theorem ;  the  square  described  on  the  hypote- 
nuse of  a  right-angled  triangle,  etc. 

Scholium. — Hence,  to  find  the  hypotenuse  of  a  right-angled  triangle, 
extract  the  square  root  of  the  sum  of  the  squares  of  the  two  sides  about 
the  right  angle, 

THEOREM    XL. 

In  any  obtuse-angled  triangle,  the  square  on  the  side  oppo- 
site the  obtuse  angle  is  greater  than  the  sum  of  the  squares 
on  the  other  two  sides,  by  twice  the  rectangle  contained  by 
either  side  about  the  obtuse  angle,  and  the  part  of  this  side 
produced  to  meet  the  perpendicular  drawn  to  it  from  the 
vertex  of  the  opposite  angle. 

Let  ABO  be  any  triangle  in  which 
the  angle  at  B  is  obtuse.  Produce 
either  side  about  the  obtuse  angle, 
as  OB,  and  from  A  draw  AD  perpen- 
dicular to  OB,  meeting  it  produced 
atD. 

It  is  obvious  that  OD  m  OB  -f  BD, 
By  Th.  36  we  have,  CD*  =  CB*  +  20B  x  BD  +  JU)', 
Adding  AD  to  each  member  of  this  equation,  we  have 

AD*  +  CD*=CB*t  BD*  +  AD1  +  20B  x  BD. 
But,  (Th.  39),  the  first  member  of  the  last  equation  is 
equal  to  AO%,  and 


BD*  -f  AD*  -  AB%, 


5* 


54 


GEOMETRY. 


Therefore,  this  equation  becomes 

~AO%  =  CB2+  AB*  +  20B  x  BD. 

That  is,  the  square  on  A  0  is  equivalent  to  the  sum  of 
the  squares  on  OB  and  AB,  increased  by  twice  the  rect- 
angle contained  by  CB  and  BD. 

Hence  the  theorem;  in  any  obtuse- angled  triangle,  the 
square  on  the  side  opposite  the  obtuse  angle,  etc. 

Scholium. — Conceive  AB  to  turn  about  the  point  A,  its  intersection 
with  CD  gradually  approaching  D.  The  last  equation  above  will  be 
true,  however  near  this  intersection  is  to  D,  and  when  it  falls  upon  D 
the  triangle  becomes  right-angled. 

In  this  case  the  line  BD  reduces  to  zero,  and  the  equation  becomes 
AC2=  CB2  +  AB*y  in  which  CB  and  AB  are  now  the  base  and  per- 
pendicular of  a  right-angled  triangle.  This  agrees  with  Theorem  39, 
as  it  should,  since  we  used  the  property  of  the  right-angled  triangle 
established  in  Theorem  39  to  demonstrate  this  proposition ;  and  in  the* 
equation  which  expresses  a  property  of  the  obtuse-angled  triangle,  we 
have  introduced  a  supposition  which  changes  it  into  one  which  is 
right-angle  1. 


THEOREM   XLI. 

In  any  triangle,  the  square  on  a  side  opposite  an  acute  angle 
is  less  than  the  sum  of  the  squares  on  the  other  two  sides,  by 
twice  the  rectangle  contained  by  either  of  these  sides,  and  the 
distance  from  the  vertex  of  the  acute  angle  to  the  foot  of  the 
perpendicular  let  fall  on  this  side,  or  side  produced,  from  the 
vertex  of  its  opposite  angle. 

Let  ABO,  either 
figure,  represent 
any  triangle ;  0  an 
acute  angle,  OB 
the  base,  and  AB 
the  perpendicular, 
which  falls  either 
without  or  on  the  base.     Now  we  are  to  prove  that 

AB*=OB2-\-  AO2— 20Bx  OB. 


BOOK   I. 


55 


From  the  firs t  figure  we  get BD=CD—CB        ( 1 ) 
and  from  the  second  BD=CB—CD        ( 2 ) 

Either  one  of  these  equations  will  give,  (Th.  37), 

BD2  =  CD2  +  CB2—2CD  x  OB. 

Adding  AD2  to  each  member  and  reducing,  we  obtaiu, 
(Th.  39),  AB2  -  AC2  +  CB2  —  2CBx  CD,  whi«h  proves 
the  proposition. 


Hence  the  theorem. 


THEOREM    XLII. 

If  in  any  triangle  a  line  be  drawn  from  any  angle  to  the 
middle  of  the  opposite  side,  twice  the  square  of  this  line, 
together  with  twice  the  square  of  one  half  the  side  bisected,  will 
be  equivalent  to  the  sum  of  the  squares  of  the  other  two  sides 

Let  AB  0  be  a  triangle,  and 
M  the  middle  point  of  its 
base. 

Then  we  are  to  prove  that 
2  AM2  +  2 CM2  =  AC2+AB*. 

Draw  AD  perpendicular  to 
the  base,  and  make  AD  =  p, 
AC=b,   AB  =  c,    CB=2a, 

AM  mm  m,  and  MD  —  x;  then  CM  =  a,  CD  =  a+x,  DB 
=  a  —  x, 

Now  by,  (Th.  39),  we  have  the  two  following  equations : 

^  +  (a  —  aO»  =  c»  (1) 

f  +  (a  +  xf  =  b*  (2) 

By  addition,  2p%  +  2x2  +  2a2  =  #•  +  <?.    But  p*  +  z»  -  m\ 
Therefore,  2m*  +  2a7  =  b*  +  <?\ 

This  equation  is  the  algebraic  enunciation  of  tho 
theorem. 


56  GEOMETRY. 

THEOREM    XLIII. 

The  two  diagonals  of  any  parallelogram  bisect  each  other ; 
and  the  sum  of  their  squares  is  equivalent  to  the  sum  of  the 
squares  of  the  four  sides  of  the  parallelogram, 

~LetABQB  be  any  parallelogram, 
and  A 0  and  BB  its  diagonals. 

We  are  now  to  prove, 

1st.  That  AM  -  EQ,  and  BE  - 
EB. 

2d.  That  AC2  +  BD2  =  ~AB2  +  ~BQ*  4-  CD2  +~AD\ 

1.  The  two  triangles  ABE  and  QBE  are  equal,  be- 
cause AB  —  QD,  the  angle  ABE  =  the  alternate  angle 
QBE,  and  the  vertical  angles  at  E  are  equal ;  therefore, 
AE,  the  side  opposite  the  angle  ABE,  is  equal  to  QE, 
the  side  opposite  the  equal  angle  QBE;  also  EB,  the 
remaining  side  of  the  one  A,  is  equal  to  EB,  the  remain- 
ing side  of  the  other  triangle. 

2.  As  AQB  is  a  triangle  whose  base,  A Q,  is  bisected 
in  E,  we  have,  by  (Th.  42), 


2AE2  +  2EB2  =  AB2  +  DC2     ( l ) 

And  as  AQB  is  a  triangle  whose  base,  ^4  C\  is  bisected 
in  i£,  we  have 

2AE2  +  2EB2==~AB2  +  BQ2     (2) 
By  adding  equations  (1)  and  (2),  and  observing  that 
EB*  m  EB2,  we  have 
4AE2  +  4EB2  -  AD2  +  2><7*  +  "AS2  +7RP 
But,  four  times  the  square  of  the  half  of  a  line  is  equiv- 
alent to  the  square  of  the  whole  line,  (Th.  36,  Corollary) ; 
therefore  ±AW  =  AQ\  and  4EB2  =  BB2;  and  by  sub- 
stituting these  values,  we  have 

AQ2  +  BB2  =~AB2  +~BQ%  +~BQ2  +  ~AB2, 
which   equation    conforms  to   the    enunciation   of   tfc* 
theorem. 


K                 H 

L 

BOOK   I.  57 

THEOREM   XLIV. 

If  a  line  be  "bisected  and  produced,  the  rectangle  contained 
by  the  whole  line  and  the  part  produced,  together  with  the 
square  of  one  half  the  bisected  line,  will  be  equivalent  to  the 
square  on  a  line  made  up  of  tlie  part  produced  and  one  half  the 
bisected  line. 

Let  AB  be  any  line,  bi- 
sected in  0  and  produced 
to  D.  On  OB  describe 
the  square  OF,  and  on 
BB  describe  the  square 
BE. 

The  sides  of  the  square 
BE  being  produced,  the  G~~  m" 

square  GrL  will  be  form- 
ed.   Also,  complete  the  construction  of  the  rectangle 
ABEK. 

Then  we  are  to  prove  that  the  rectangle,  AE,  and  the 
aquare,  GrL,  are  together  equivalent  to  the  square, 
CBFa. 

The  two  complementary  rectangles,  CL  and  LF,  are 
equal,  (Th.  31).  But  OL=AH,  the  line  AB  being  bisected 
at  0;  therefore  AL  is  equal  to  the  sum  of  the  two  com- 
plementary rectangles  of  the  square  OF.  To  AL  add 
the  square  BE,  and  the  whole  rectangle,  AE,  will  be 
equal  to  the  two  rectangles  CE  and  EM.  To  each  of 
these  equals  add  EM,  or  the  square  on  HL  or  its  equal 
OB,  and  we  have  rectangle  AE  -f  square  EM  =  OB2 ; 
but  rectangle  AE  =  AB  x  BB,  and  square  EM  =  CB2* 
Hence  the  theorem,  etc. 

Scholium,  —  If  we  represent  AB  by  2a,  and  BD  by  x,  then  AD  — 
2a  -f  x,  and  AD  X  BD  =  2ax  +  x*.  But  CB1  =  a' ;  adding  this 
equation  to  the  preceding,  member  to  member,  we  get  AD  X  BD  -f- 
CB2  =  a* -f-  2ax  -f  x3  =  a  +  x\  But  CD  — a  +  s;  hence  this  equa- 
tion is  equivalent  to  the  equation  AL  X  DB  -f-  CE?  =  CD  ,  which  is 
the  algebraic  proof  of  the  theorem. 


58  GEOMETRY 

THEOREM  XLV. 

If  a  straight  line  be  divided  into  two  equal  parts,  and  also 
into  two  unequal  parts,  the  rectangle  contained  by  the  two  un- 
equal parts  together  with  the  square  of  the  line  between  the 
points  of  division,  will  be  equivalent  to  the  square  on  one  half 
the  line. 

Let  AB  be  a  line  bisected  in  0,  and  divided  into  two 
unequal  parts  in  B, 

We  are  to  prove 

that  AB  x  BB  +         . p c 

Cff^AQ\ov~CB\        B  A 

We  see  by  inspection  that  AB  =  AC  +  CB,  and  BB 
m,  AC—  OB;  therefore  by  (Th.  38),  we  have 
ABx  BB  =  AC2—~CB\ 

By  adding  UB2  to  each  of  these  equals,  we  obtain 
AB  x  BB  +  ~CD*  =~AC% 

Hence  the  theorem. 


BOOK  II.  59 


BOOK  II 


PROPORTION. 
DEFINITIONS  AND  EXPLANATIONS. 

The  word  Proportion,  in  its  common  meaning,  de- 
notes that  general  relation  or  symmetry  existing  between 
the  different  parts  of  an  object  which  renders  it  agree- 
able to  our  taste,  and  conformable  to  our  ideas  of  beauty 
or  utility ;  but  in  a  mathematical  sense. 

1.  Proportion  is  the  numerical  relation  which  one  quan- 
tity bears  to  another  of  the  same  kind. 

As  the  magnitudes  compared  must  be  of  the  same  kind, 
proportion  in  geometry  can  be  only  that  of  a  line  to  a 
line,  a  surface  to  a  surface,  an  angle  to  an  angle,  or  a  volume 
to  a  volume. 

2.  Ratio  is  a  term  by  which  the  number  which  meas- 
ures the  proportion  between  two  magnitudes  is  desig- 
nated, and  is  the  quotient  obtained  by  dividing  the  one 

by  the  other.     Thus,  the  ratio  of  A  to  B  is  --,  or  A  :  B, 

in  which  A  is  called  the  antecedent,  and  B  the  consequent. 
If,  therefore,  the  magnitude  A  be  assumed  as  the  unit  or 
standard,  this  quotient  is  the  numerical  value  of  B  ex- 
pressed in  terms  of  this  unit. 

It  is  to  be  remarked  that  this  principle  lies  at  the  found- 
ation of  the  method  of  representing  quantities  by  num- 
bers. For  example,  when  we  say  that  a  body  weighs 
twenty-five  pounds,  it  is  implied  that  the  weight  of  this 
body  has  been  compared,  directly  or  indirectly,  with  that 
of  the   standard,  one  pound.     And  so   of  geometrica 


60  GEOMETRY. 

magnitudes ;  when  a  line,  a  surface,  or  a  volume  is  said 
to  be  fifteen  linear,  superficial,  or  cubical  feet,  it  is  un- 
derstood that  it  has  been  referred  to  its  particular  unit, 
and  found  to  contain  it  fifteen  times ;  that  is,  fifteen  is 
the  ratio  of  the  unit  to  the  magnitude. 

"When  two  magnitudes  are  referred  to  the  same  unit, 
the  ratio  of  the  numbers  expressing  them  will  be  the 
ratio  of  the  magnitudes  themselves. 

Thus,  if  A  and  B  have  a  common  unit,  a,  which  is 
contained  in  A,  m  times,  and  in  B,  n  times,  then  A  =•  ma 

j  t>  ,  B       na       n 

and  B  =  na.  and  -r  =  —  =  — . 
A      ma      m 

To   illustrate,  let  the 

line  A  contain  the  line       .  A 

a  six  times,  and  let  the  , 

line  B  contain  the  same  a 

line  a  five  times :   then 

i        i        i j j j 

A= 6a  and  B—5a,  which  B 

.      B      ba      5 
glVe2=6-a  =  6- 

3.  A  Proportion  is  a  formal  statement  of  the  equality 
of  two  ratios. 

Thus,  if  we  have  the  four  magnitudes  A,  B,  0  and  D, 

fmch  that  -r  =  ^,  this  relation  is  expressed  by  the  pro- 
portion A:  B  ::  C:  D,  or  A  :  B  =  0 :  D,  the  first  of 
which  is  read,  A  is  to  B  as  O  is  to  D ;  and  the  second, 
the  ratio  of  A  to  B  is  equal  to  that  of  0  to  D. 

4.  The  Terms  of  a  proportion  are  the  magnitudes,  or 
caore  properly  the  representatives  of  the  magnitudes 
compared. 

5.  The  Extremes  of  a  proport  on  are  its  first  and  fourth 
terms. 

6.  The  Means  of  a  proportion  are  its  second  and  third 
terms. 

7.  A  Conplet  consists  of  the  two  terms  of  a  ratio.    The 


BOOK   II.  61 

first  and  second  terms  of  a  proportion  are  called  the 
first  couplet,  and  the  third  and  fourth  terms  are  called 
the  second  couplet. 

8.  The  Antecedents  of  a  proportion  are  its  first  and 
third  terms. 

9.  The  Consequents  of  a  proportion  are  its  second  and 
fourth  terms. 

In  expressing  the  equality  of  ratios  in  the  form  of  a 
proportion,  we  may  make  the  denominators  the  ante- 
cedents, and  the  numerators  the  consequents,  or  the 
reverse,  without  affecting  the  relation  "between  the  magni- 
tudes. It  is,  however,  a  matter  of  some  little  importance 
to  the  "beginner  to  adopt  a  uniform  rule  for  writing  the 
terms  of  the  ratios  in  the  proportion ;  and  we  shall  always, 
unless  otherwise  stated,  make  the  denominators  of  the 
ratios  the  antecedents,  and  the  numerators  the  conse- 
quents.* 

10.  Equimultiples  of  magnitudes  are  the  products  arising 
from  multiplying  the  magnitudes  by  the  same  number. 
Thus,  the  products,  Am  and  Bm,  are  equimultiples  of 
A  and  B. 

U.  A  Mean  Proportional  between  two  magnitudes  is  a 
magnitude  which  will  form  with  the  two  a  proportion, 
when  it  is  made  a  consequent  in  the  first  ratio,  and  an 
antecedent  in  the  second.  Thus,  if  we  have  three  mag- 
nitudes A,  By  and  0,  such  that  A  :  B  : :  B  :  (7,  B  is  a 
mean  proportional  between  A  and  0. 

12.  Two  magnitudes  are  reciprocally,  or  inversely  pro- 
portional when,  in  undergoing  changes  in  value,  one  is 
multiplied  and  the  other  is  divided  by  the  same  number. 

Thus,  if  A  and  B  be  two  magnitudes,  so  related  that  when 

j> 
A  becomes  mA,  B  becomes  — ,  A  and  B  are  said  to  be 

m 

inversely  proportional. 

*  For  discussion  of  the  two  methods  of  expressing  Ratio,  see  Uni 
vorsity  Algebra. 
6 


62  GEOMETRY. 

13.  A  Proportion  is  taken  inversely  when  the  ante- 
cedents are  made  the  consequents  and  the  consequents 
the  antecedents. 

14.  A  Proportion  is  taken  alternately,  or  by  alternation, 
when  the  antecedents  are  made  one  couplet  and  the  con- 
sequents the  other. 

15.  Mutually  Equiangular  Polygons  have  the  same  num- 
ber of  angles,  those  of  the  one  equal  to  those  of  the 
ethers,  each  to  each,  and  the  angles  like  placed. 

16.  Similar  Polygons  are  such  as  are  mutually  equi- 
angular, and  have  the  sides  about  the  equal  angles,  taken 
in  the  same  order,  proportional. 

17.  Homologous  Angles  in  similar  polygons  are  those 
which  are  equal  and  like  placed ;  and 

18.  The  Homologous  Sides  are  those  which  are  like  dis- 
posed about  the  homologous  angles. 

THEOREM  I. 

If  the  first  and  second  of  four  magnitudes  are  equal,  and 
also  the  third  and  fourth,  the  four  magnitudes  may  form  a 
proportion. 

Let  A,  B,  0,  and  B  represent  four  magnitudes,  such 
that  A  —  B  and  C—B;  we  are  to  prove  that  A  :  B  : : 
C  :  B. 

Now,  by  hypothesis,  A  is  equal  to  B,  and  their  ratio  is 
therefore  1 ;  and  since,  by  hypothesis,  0  is  equal  to  B, 
their  ratio  is  also  1. 

Hence,  the  ratio  of  A  to  B  is  equal  to  that  of  C  to  B ; 
and,  (by  Def.  3), 

A  :  B  : :  0  :  D. 

Therefore,  four  magnitudes  which  are  equal,  two  and 
two,  constitute  a  proportion. 


BOOK  II.  63 


THEOREM    II. 

If  four  magnitudes  constitute  a  proportion,  the  product  of 
the  extremes  is  equal  to  the  product  of  the  means. 

Let  the  four  magnitudes  A,  B,  C,  and  D  form  the  pro- 
portion A  :  B  : :  0  :  D ;  we  are  to  prove  that  A  x  D 
=  Bx  C. 

The  ratio  of  A  to  B  is  expressed  hy  -j  =  r. 
The  ratio  of  0  to  B  is  expressed  hy  ^  =  r. 

Hence,  (Ax.  1),  ^  -  ~. 

Multiplying  each  of  these  equals  hy  A  x  C,  we  have 
B  x  C=Ax  D. 

Hence  the  theorem ;  if  four  magnitudes  are  in  propor- 
Hon,  etc. 

Cor.  1.  Conversely;  If  we  ha/ve  the  product  of  two  mag- 
nitudes equal  to  the  product  of  two  other  magnitudes,  they 
will  constitute  a  proportion  of  which  either  two  may  be 
made  the  extremes  and  the  other  two  the  means. 

Let  the  magnitudes  B  x  C  =  A  x  D.    Dividing  both 
members     of    the     equation     by     A  x  C,    we     obtain 
B  _B 
A~  C 

Hence  the  proportion  A  :  B  : :  (7:2). 

Cor.  2.  If  we  divide  both  members  of  the  equation 
Ax  B  =  B  x  C    by  A, 

we  have         D  =  — -A — . 
A 

That  is,  to  find  the  fourth  term  of  a  proportion,  mul- 
tiply the  second  and  third  terms  together  and  divide  the  pro- 
duct by  the  first  term.  This  is  the  Rule  of  Three  of 
Arithmetic- 


64  .  GEOMETRY. 

This  equation  shows  that  any  one  of  the  foui  terms 
can  be  found  by  a  like  process,  provided  the  other  three 
are  given. 

THEOREM    III. 

If  three  magnitudes  are  continued  proportionals,  the  product 
of  the  extremes  is  equal  to  the  square  of  the  mean. 

Let  A,  B,  and  0  represent  the  three  magnitudes : 

Then  A  :  B  : :  B  :  C,  (by  Def.  11). 

But,  (by  Th.  2),  the  product  of  the  extremes  is  equal 
to  the  product  of  the  means ;  that  is,  A  x  (7=  B*. 

Hence  the  theorem ;  if  three  magnitudes,  etc, 

THEOREM   IV. 

Equimultiples  of  any  two  magnitudes  have  the  same  ratio 
as  the  magnitudes  themselves  ;  and  the  magnitudes  and  their 
equimultiples  may  therefore  form  a  proportion. 

Let  A  and  B  represent  two  magnitudes,  and  mA  and 
mB  their  equimultiples. 

Then  we  are  to  prove  that  A  :  B  : :  mA  •  *nB, 

The   ratio  of  A  to  B  is  ~,  and  of  mA  Us  mB  is 

mB      B  '.  ,. 

— T  =  -r,  the  same  ratio. 
mA      A7 

Hence  the  theorem;  equimultiples  of  any  twc  &4y*i 

tudes,  etc, 

THEOREM    V. 

If  four  magnitudes  are  proportional,  they  will  be  propor~ 
tional  when  taken  inversely. 

If  A  :  B  : :  mA  :  mB,  then  B  :  A  : :  mB  :  mA ; 

For  in  either  case,  the  product  of  the  extremes  equals 
that  of  the  means;  or  the  ratio  of  the  couplets  is  the 
same. 

Hence  the  theorem ;  if  four  quantities  are  propor- 
tional, etc. 


BOOK   II 


THEOREM   VI. 


Magnitudes  which  are  proportional  to  the  same  propor 
tionals,  are  proportional  to  each  other. 
If     A  :  B  =  P  :  Q  1      Then  we  are  to  prove  that 
and    a  :    b  =  P  :  Q)  A  :  B  =  a  :  b. 

From  the  1st  proportion,  —  =  ~ ; 

A.        Jr 

From  the  2d  "  -  =  -^; 

a      P 

7?       h 

Therefore,  by  (Ax.  1),      -j  =  -,  or  A  :  B  =  a  :  b. 

jA.      a 

Hence  the  theorem ;  magnitudes  which  are  proportional 
to  the  same  proportionals,  etc. 

Cor.  1.  This  principle  may  be  extended  through  any 
number  of  proportionals. 

Cor.  2.  If  the  ratio  of  an  antecedent  and  consequent  of  one 
proportion  is  equal  to  the  ratio  of  an  antecedent  and  conse- 
quent of  another  proportion,  the  remaining  terms  of  the  two 
proportions  are  proportional. 

For,  if  A  :  B  : :  C  :  D 

and  M  :  N  : :  P  :  Q 

in  which  A=WthenC=P> 

hence  0  :  D  : :  P  :  Q. 

THEOREM   VII. 

If  any  number  of  magnitudes  are  proportional,  any  one  of 
the  antecedents  will  be  to  its  consequent  as  the  sum  of  all  the 
antecedents  is  to  the  sum  of  all  the  consequents. 

Let  A,  B,  C,  B,  E,  etc.,  represent  the  several  magni 
tudes  whi  3h  give  the  proportions 


6* 


A  :  B 

::  0  :  D 

A  :  B 

::  E  :  F 

A  :  B 

: :   a  :  H, 

E 

etc.,  etc 


66  GEOMETRY. 

To  which  we  may  annex  the  identical  proportion, 

A  :  B  : :  A  :  B. 
Now,  (by  Th.  2),  these  proportions  give  the  following 
equations, 

A  x  D  -  B  x   0 
A  x   F  =  B  x  E 
A  x  H=  B  x    a 
A  x  B  =  B  x  A,  etc.  etc. 
From  which,  by  addition,  there  results  the  equation, 
A(B  +  D  +  F  +  H,  etc.)  =  B(A+  C+F+  #,  etc.) 
But  the  sums  B  +  D  +  F,  etc.,  and  A  -f  C  +  U,  etc., 
may  be  separately  regarded  as  single  magnitudes ;  there- 
fore, (Th.  2,  Cor.  1), 

A  :  B  ::  A+C+F+  G,  etc.  :  .B  +  D  -f  #+  J7",  etc. 

Hence  the  theorem ;  if  any  number  of  magnitudes  are  pro- 
portional,  etc. 

THEOREM   VIII. 

If  four  magnitudes  constitute  a  proportion,  the  first  will  be 
to  the  sum  of  the  first  and  second  as  the  third  is  to  the  sum  of 
the  third  and  fourth. 

By  hypothesis,  A  :  B  ::  C :  D;  then  we  are  to  prove 
that  A  :  A  +  B  ::  0  :  C  +  D. 

By  the  given  proportion,  —  =  — . 
Adding  unity  to  both  members,  and  reducing  them  to 
the  form  of  a  fraction,  we  have  — - —  =  — ^ — .   Chang- 
ing this  equation  into  its  equivalent  proportion al  form, 
we  have 

A  :  A  4-  B  : :  0  :  C  +  B. 

Hence  the  theorem  ;  if  four  magnitudes  constitute  a  pro* 
portion,  etc. 

T> 

Cor.  If  we  subtract  each  member  of  the  equation  -j  = 


BOOK   II.  67 

D 

p  from  unity,  and  reduce  as  before,  we  shall  have 

A  :  A  —  B  ::  0  :  C  —  D. 
Hence  also ;  if  four  magnitudes  constitute  a  proportion, 
the  first  is  to  the  difference  between  the  first  and  second,  as  the 
third  is  to  the  difference  between  the  third  and  fourth. 

THEOREM   IX. 

If  four  magnitudes  are  proportional,  the  sum  of  the  first  and 
second  is  to  their  difference  as  the  sum  of  the  third  and  fourth 
is  to  their  difference. 

Let  A,  B,  0,  and  D  be  the  four  magnitudes  which  give 
the  proportion 

A  :  B  ::  C  :  D; 

we  are  then  to  prove  that  they  will  also  give  the  propor- 
tion 

A  +  B  :  A—B  ::  C  +  D  :  Q  —  D. 
By  Th.  8  we  have  A  :  A  +  B  -  0  :  (7+2). 
Aisoby  Corollary,  sameTh.,  A  :  A  —  B  =  C :  C—D. 
Xow,  if  we  change  the  order  of  the  means  in  these  pro- 
portions, which  may  be  done,  since  the  products  of  ex- 
tremes and  means  remain  the  same,  we  shall  have 
A  :  C  =  A  +  B  :  C+  B. 
A  :  C  =  A  —  B  :  C—B. 
Hence,  (Th.  6),  we  have 

A  +  B  :  0+  D  -  A  —  B  :  C—D. 
Or,       A  +  B  :  A  —  B  -  0  +  D  :  Q—D. 
Ilence  the  theorem ;  if  four  magnitudes  are  proportional, 
etc. 

THEOREM   X. 

If  four  magnitudes  are  proportional,  like  powers  or  like 
roots  of  the  same  magnitudes  are  also  proportional. 

If  the  four  magnitudes,  A,  B,  C,  and  D,  give  the  pro- 
portion 


68  GEOMETRY. 

A  :  B  : :  0  :  D, 

we  are  to  prove  that 

An  :  Bn  ::  Cn  :  Bn. 

The  hypothesis  gives  the  equation  —  =  — .     Eaising 

A        G 

both  members  of  this  equation  to  the  nth  power,  we  have 

Bn     Dn 

-^  =  -^,  which,  expressed  in  its  equivalent  proportional 

form,  gives 

An  :  Bn  ::  Cn  :  Bn. 

If  n  is  a  whole  number,  the  terms  of  the  given  propor- 
tion are  each  raised  to  a  power ;  but  if  n  is  a  fraction 
having  unity  for  its  numerator,  and  a  whole  number  for  its 
denominator,  like  roots  of  each  are  taken. 

As  the  terms  of  the  proportion  may  be  first  raised  to 
like  powers,  and  then  like  roots  of  the  resulting  propor- 
tion be  taken,  n  may  be  any  number  whatever. 

Hence  the  theorem ;  if  four  magnitudes,  etc. 

THEOREM    XI. 

If  four  magnitudes  are  proportional,  and  also  four  others, 
the  products  which  arise  from  multiplying  the  first  four  by  the 
second  four,  term  by  term,  are  also  proportional. 


Admitting  that  A  :     B 

and  X:    Y 


We  are  to  show  that  AX :  BY 


0  :  B, 
Mi  N, 


QMiBK 


From  the  first  proportion,  —  =  —  ; 

a.      o 

Y     N 
From  the  second,  __  =  -_. 

X     M 

Multiply  these  equations,  member  by  member,  and 

AX     CM' 
Or,  AX  i  BY  n  CM:  BN. 

The  same  would  be  true  in  any  number  oi  proportions. 
Hence  the  theorem ;  if  four  magnitudes  are,  etc. 


BOOK   II.  69 

THEOREM   XII. 

If  four  magnitude*  are  'proportional,  and  also  four  others, 
the  quotients  which  arise  from  dividing  the  first  four  by  the 
swond  four,  term  by  term,  are  proportional. 
By  hypothesis,        A  :  B  : :  0  :  D, 
and  X  :  Y  ::M:  K 

Multiply  extremes  and  means,     AD  ==  OB,       ( 1 ) 
and  XN=MY.      (2) 

Divide  (1)  hy  (2),  and     ±  x  *  ?  JJ  x  * 

Convert  these  four  factors,  which  make  two  equal  pro- 
ducts, into  a  proportion,  and  we  have 

X  :   Y':  Ml  JV" 

By  comparing  this  with  the  given  proportions,  we  find 
it  is  composed  of  the  quotients  of  the  several  terms  of 
the  first  proportion,  divided  by  the  corresponding  terms 
of  the  second. 

Hence  the  theorem ;  if  four  magnitudes  are  proportional, 
etc. 

THEOREM    XIII. 

If  four  magnitudes  are  proportional,  we  may  multiply  the 
first  couplet,  the  second  couplet,  the  antecedents  or  the  conse- 
quents, or  divide  them  by  the  same  quantity,  and  the  results 
will  be  proportional  in  every  case. 

Let  the  four  magnitudes  A,  B,  0,  and  D  give  the  pro- 
portion A:  B  ::  0:  D.  By  multiplying  the  extremes 
and  means  we  have 

A.D  =  B.O         (1) 

Multiply  both  members  of  this  equation  by  any  num- 
ber, as  a,  and  we  have 

aA.D  =  aB.O 

By  converting  this  equation  into  a  proportion  in  four 
different  ways,  we  have  as  follows  : 


70 


GEOMETRK 

aA 

:  aB 

::  0 

:  B 

A  : 

B  :: 

aO  : 

aB 

aA 

:  B  : 

:  aQ 

:  B 

A  : 

aB  : 

:  C: 

aB. 

Kesuming^the  original  equation,  (1),  and  dividing  both 
members  by  a,  we  have 

A.B  _  B.Q 
a  a 

This  equation  may  also  be  converted  into  a  proportion 
in  four  different  ways,  with  the  following  results : 


; 

x> 

:  : 

0  : 

D 

a 

a 

A  : 

B 

•  : 

0  . 
a 

B 

a 

A  . 

B 

. . 

(7. 

B 

a 

a 

A  : 

B 
a 

: : 

C  : 

B 

a 

Hence  the  theorem ;  if  four  magnitudes  are  in  proportion, 
etc, 

THEOREM    XIV. 

If  three  magnitudes  are  in  proportion,  the  first  is  to  the 
third  as  the  square  of  the  first  is  to  the  square  of  the  second. 

Let  A,  B,  and  C,  be  three  proportionals. 

Then  we  are  to  prove  that  A  :  C—  A2 :  B* 

By(Th.  3)  AC=B> 

Multiply  this  equation  by  the  numeral  value  of  A,  and 
we  have  %  A*C=A& 

This  equation  gives  the  following  proportion : 

Ax  C=A2:B\ 

Hence  the  theorem. 

Remark.  —  It  is  now  proposed  to  make  an  application  of  the  pre- 
ceding abstract  principles  of  proportion,  in  geometrical  investigations 


BOOK    II 


71 


D 


THEOREM  XV. 
If  two  parallelograms  are  equal  in  area,  the  base  and  per- 
pendicular of  either  may  be  made  the  extremes  of  a  propor- 
tion, of  which  the  base  and  perpendicular  of  the  other  are  the 
means. 

Let  ABQD, 
and  HLNM, 
be  two  paral- 
lelograms hav- 
ing equal  areas,  A 
by  hypothesis ;  then  we  are  to  prove  that 

AB  :  LN  : :  MK  :  BF, 
in  which  MK  and  BF  are  the 
altitudes  or  perpendiculars  of 
the  parallelograms. 

This  proportion  is  true,  if 
the  product  of  the  extremes 
is  equal. to  the  product  of  the  means; 
that  is,  if  the  equation 

AB.BF  =  LN.MKis  true. 
But  AB.BF  is  the  measure  of  the  rectangle  ABFE, 
by  (Definition  54,  B.  I.),  and  this  rectangle  is  equal  in 
area  to  the  parallelogram  ABOD,  (B.  I.,  Th.  27). 

In  the  same  manner,  we  may  prove  that  LN.MK  is 
the  measure  of  the  parallelogram  NLHM.  But  these 
two  parallelograms  have  equal  areas  by  hypothesis. 

Therefore,  AB.BF  =  LN.MK  is  a  true  equation,  and 
Th.  2,  Cor.  1),  gives  the  proportion 

AB  :  LN  : :  MK  :  BF. 
Hence  the  theorem ;  if  two  parallelograms  are  equal  in 
area,  etc. 

THEOREM  XVI. 
Parallelograms  having  equal  altitudes  are  to  each  other  as 
their  bases. 

Since  parallelograms  having  equal  bases  and  equal 
altitudes  are  equal  in  area,  however  much  their  angle* 


72  GEOMETRY. 

may  differ,  we  can  suppose  the  two  parallelograms  under 
consideration  to  be  mutually  equiangular,  without  in  the 
least  impairing  the  generality  of  this  theorem.  There- 
fore, let  ABOB 
and  AEFB  be  two 
parallelograms 
having  equal  alti- 
tudes, and  let  them 
be  placed  with 
their  bases  on  the  same  line  AE,  and  let  the  side,  ABt 
be  common.  First  suppose  their  bases  commensurable, 
and  that  AE  being  divided  into  nine  equal  parts,  AB 
contains  five  of  those  parts. 

If,  through  the  points  of  division,  lines  be  drawn  paral- 
lel to  AB,  it  is  obvious  that  the  whole  figure,  or  the 
parallelogram,  AEFB,  will  be  divided  into  nine  equal 
parts,  and  that  the  parallelogram,  ABOB,  will  be  com- 
posed of  five  of  those  parts. 

Therefore,  ABCB  :  AEFB  : :  AB  :  AE  : :  *5  :  9. 

Whatever  be  the  whole  numbers  having  to  each  other 
the  ratio  of  the  lines  AB  and  AE,  the  reasoning  would 
remain  the  same,  and  the  proportion  is  established  when 
the  bases  are  commensurable.  But  if  the  bases  are  not 
to  each  other  in  the  ratio  of  any  two  whole  numbers,  it 
remains  still  to  be  shown  that 

AEFB  :  ABOB  ::  AE  :  AB    (1) 

If  this  propor- 
tion  is  not  true, 
there  must  be  a 
line  greater  or  less 
than  AB,  to  which 

AE  will  have  the    A  B  L 

same  ratio  that  AEFB  has  to  ABOB. 

Suppose  the  fourth  proportional  greater  than  AB,  as 
AK,  then, 

AEFB  :  ABOB  ::  AE  :  AK    (2). 


BOOK    II.  78 

If  we  now  divide  the  line  AE  into  equal  parts,  each 
less  than  the  line  BK,  one  point  of  division,  at  least,  will 
fall  between  B  and  K.  Let  L  be  such  point,  and  draw 
LM  parallel  to  BO. 

This  construction  makes  AE  and  AL  commensura- 
ble; and  by  what  has  been  already  demonstrated,  we 

have 

AEFD  :  ALMD  ::  AE  :  AL.     (3) 

Inverting  the  means  in  proportions  ( 2 )  and  ( 3 ),  they 

become 

AEFD  :  AE  ::  ABQD  :  AK; 

and  AEFD  :  AE  : :  ALMD  :  AL. 

Hence,  (Th.  6), 

ABOD  :  AK  : :  ALMD  :  AL. 
By  inverting  the  means  in  this  last  proportion,  we  have 

ABOD  :  ALMD  : :  AK  :  AL. 

But  AK  is,  by  hypothesis,  greater  than  AL;  hence,  if 
this  proportion  is  true,  ABOD  must  be  greater  than 
ALMD;  but  on  the  contrary  it  is  less.  We  therefore 
conclude  that  the  supposition,  that  the  fourth  propor- 
tional, AK,  is  greater  than  AB,  from  which  alone  this 
absurd  proportion  results,  is  itself  absurd. 

In  a  similar  manner  it  can  be  proved  absurd  to  sup- 
pose the  fourth  proportional  less  than  AB. 

Therefore  the  fourth  term  of  the  proportion  ( 1 )  can  be 
neither  less  nor  greater  than  AB ;  it  is  then  AB  itself, 
and  parallelograms  having  equal  altitudes  are  to  each 
other  as  their  bases,  whether  these  bases  are  commensur- 
able or  not. 

Hence  the  theorem ;  Parallelograms  having  equal  alti- 
tudes, etc. 

Oor.  1.  Since  a  triangle  is  one  half  of  a  parallelogram 
having  the  same  base  as  the  triangle  and  an  equal  alti- 
tude, and  as  the  halves  of  magnitudes  have  the  same 
ratio  as  their  wholes ;  therefore, 
7 


74 


GEOMETRY. 


Triangles  having  the  same  or  equal  altitudes  are  to  each 
other  as  their  bases. 

Cor.  2.  Any  triangle  has  the  same  area  as  a  right- 
angled  triangle  having  the  same  hase  and  an  equal  alti- 
tude ;  and  as  either  side  about  the  right  angle  of  aright- 
angled  triangle  may  be  taken  as  the  base,  it  follows  that 

Two  triangles  having  the  same  or  equal  bases  are  to  each 
other  as  their  altitudes. 

Cor.  3.  Since  either  side  of  a  parallelogram  may  be 
taken  as  its  base,  it  follows  from  this  theorem  that 

Parallelograms  having  equal  bases  are  to  each  other  as  their 
altitudes. 


THEOREM   XVII. 

If  lines  are  drawn  cutting  the  sides,  or  the  sides  'produced,  of 
a  triangle  proportionally,  such  secant  lines  are  parallel  to  the 
base  of  the  triangle ;  and  conversely,  lines  drawn  parallel 
to  the  base  of  a  triangle  cut  the  sides,  or  the  sides  produced, 
proportionally. 

Let  ABC  be  any  triangle,  and 
draw  the  line  BE  dividing  the  sides 
AB  and  AC  into  parts  wmich  give 
the  proportion 

AB  :  BB  : :  AE  :  EC. 
We  are  to  prove  that  BE  is  parallel 
to  BC. 

If  BE  is  not  a  parallel  through 
the  point  B  to  the  line  BC,  suppose 
Bm  to  be  that  parallel ;  and  draw  the 
lines  BC  and  Bm. 

Now,  the  two  triangles  ABm  and 
mBC,  have  the  same  altitude,  since 
they  have  a  common  vertex,  B,  and  their  bases  in  the 
same  line,  AC;  hence,  they  are  to  each  other  as  their 
bases, Am  and  mC,  (Th.  16,  Cor.  1). 


BOOK   II.  75 

That  is,      A  ADm  :  A  mDQ  : :  Am  :  mO, 
Also,  A  AmD  :  £  J>mB  ::  AD  :  DB. 

But,  since  Dm  is  supposed  parallel  to  BO,  the  triangles 
DBm  and  DCm  have  equal  areas,  because  they  are  on 
the  same  base  and  between  the  same  parallels,  (Th.  28, 
B.I). 

Therefore  the  terms  of  the  first  couplets  in  the  two 
preceding  proportions  are  equal  each  to  each,  and  conse- 
quently the  terms  of  the  second  couplets  are  proportional, 
(Theorem  6). 

That  is,   AD  :  DB  : :  Am  :  mO 

But  AD  :  DB  ::  AE  :  EQ  by  hypothesis. 

Hence  we  again  have  two  proportions  having  the  first 
couplets,  the  same  in  both,  and  we  therefore  have 

AE  :  EC  : :  Am  :  mO 

By  alternation  this  becomes 

AE  :  Am  ::  EC  :  mO 

That  is,  AE  is  to  Am,  a  greater  magnitude  is  to  a  less, 
as  EQ  is  to  mO,  a  less  to  a  greater,  which  is  absurd. 
Had  we  supposed  the  point  m  to  fall  between  E  and  0, 
our  conclusion  would  have  been  equally  absurd ;  hence 
the  suppositions  which  have  led  to  these  absurd  results 
are  themselves  absurd,  and  the  line  drawn  through  the 
point  D  parallel  to  BO  must  intersect  AO  in  the  point 
E.  Therefore  the  parallel  and  the  line  DE  are  one  and 
the  same  line. 

Conversely :  JfDE  be  drawn  parallel  to  the  base  of  the 
triangle,  then  will 

AD  i  DB  ::  AE  i  EQ 
For  as  before, 

A  ADE  :  a  EDO  : :  AE  :  EQ 
and      A  DEB   :  A  ADE  x;  DB  i  AD 

Multiplying  the  corresponding  terms  of  these  propor- 


76  GEOMETRY. 

tions,  and  omitting  the  common  factor,  a  ADE,  in  the 
first  couplet,  we  have 

A  DEB  :  A  EDO  : :  AE  x  DB  :  EC  x  AD. 

But  the  a's  DEB  and  EDO  have  equal  areas,  (Th.  28, 
B.  I) ;  hence  AE  x  DB  =  EC  x  AD,  which  in  the  form 
of  a  proportion  is 

AE  :  EC  : :  AD  :  DB 

or,  AD  :  DB  : :  AE  :  EC 

and  therefore  the  line  parallel  to  the  base  of  the  triangle, 
divides  the  sides  proportionally. 

It  is  evident  that  the  reasoning  would  remain  the  same, 
had  we  conceived  ADE  to  be  the  triangle  and  the  sides 
to  be  produced  to  the  points  B  and  0. 

Hence  the  theorem;  if  lines  are  drawn  cutting  the 
sides,  etc. 

Cor.  1.  Because  DE  is  parallel  to  BO,  and  intersects 
the  sides  AB  and  A 0,  the  angles  ADE  and  ABO  are 
equal.  For  the  same  reason  the  angles  AED  and  AOB 
are  equal,  and  the  A's  ADE  and  ABO  are  equiangular. 

Let  us  now  take  up  the  triangle  ADE,  and  place  it  on 
ABO;  the  angle  ADE  falling  on  |__  B,  the  side  AD  on 
the  side  AB,  and  the  side  DE  on  the  side  BO 

Now,  since  the  angle  A  is  common,  and  the  angles 
AED  and  A  OB  are  equal,  the  side  AE  of  the  A  ADE, 
in  its  new  position,  will  be  parallel  to  the  side  AO  of  the 
A  ABO. 

The  last  proportion  of  this  Th.  gives  (Th.  8  and  Th.  5), 
AD  :  AE  w  AB  :  AC 

From  the  above  construction  we  obtain,  by  a  similar 
course  of  reasoning,  the  proportion 

AD  :  DE  : :  AB  :  BO 

And  in  like  manner  it  may  be  shown  that 
AE  :  ED  ::  AO  :  OB 

That  is,  the  sides  about  the  equal  angles  of  equiangular 
triangles,  taken  in  the  same  order,  are  proportional,  and  the 
triangles  are  similar,  (Def.  16). 


BOOK   II 


77 


Cor.  2.  Two  triangles  having  an  angle  in  one  equal  to  an 
angle  in  the  other,  and  the  sides  about  these  equal  angles  pro* 
portional,  are  equiangular  and  similar. 

For,  if  the  smaller  triangle  be  placed  on  the  larger, 
the  equal  angles  of  the  triangles  coinciding,  then  will 
the  sides  opposite  these  angles  be  parallel,  and  the  triau- 
gles  will  therefore  be  equiangular  and  similar. 


THEOREM    XVIII. 

If  any  triangle  have  its  sides  respectively  proportional  to 
the  like  or  homologous  sides  of  another  triangle,  each  to  each, 
then  the  two  triangles  will  be  equiangular  and  similar. 

Let  the  triangle  abc  have  its  sides  pro- 
portional to  the  triangle  ABO  ;  that  is,  ac 
to  A 0  as  cb  to  OB,  and  ac  to  AC  as  ab  to 
AB ;  then  we  are  to  prove  that 
the  a's,  abc  and  ABO,  are  equi- 
angular and  similar. 

On  the  other  side  of  the  base, 
AB,  and  from  A,  conceive  the 
angle  BAB  to  be  drawn  =  to  the 
|__  a ;  and  from  the  point  B, 
conceive  the  angle  ABB  to  be 
drawn  =  to  the  [_  b.  Then  the  third  [__  D  must  be  =a 
to  the  third  [_  o,  (B.  I,  Th.  12,  Cor.  2) ;  and  the  A  ABB 
will  be  equiangular  to  the  A  abc  by  construction. 

Therefore,  ac  :  ab  =  AB  :  AB 

By  hypothesis,  ac  :  ab  =  A  O  :  AB 

Hence,  AB  :  AB  =  A  0  :  A B,  (Th.  6). 

In  this  last  proportion  the  consequents  are  equal; 
therefore,  the  antecedents  are  equal :  that  is, 
AB  -  AO 

In  the  sa me  manner  we  may  prove  that 
BB  =    OB 


78  GEOMETRY. 

But  AB  is  common  to  the  two  triangles ;  therefore, 
the  three  iides  of  the  A  ABB  are  respectively  equal  to 
the  three  sides  of  the  A  ABO,  and  the  two  a's  are  equal, 
(B.  I,  Th.  21). 

But  the  A's  ABB,  and  abc,  are  equiangular  by  con- 
struction; therefore,  the  A's,  ABO,  and  abc,  are  also 
equiangular  and  similar. 

Hence  the  theorem ;  if  any  triangle  have  its  sides,  etc, 

Second  Demonstration. 

Let  abc  and  ABO  be  two  triangles 
whose  sides  are  respectively  propor- 
tional, then  will  the  triangles  be  equi- 
angular and  similar. 

That  is,  L«  =  LA  L&  =  L  2>  and 

If  the  [_  c  be  in  fact 
equal  to  the  [__  0,  the  tri- 
angle abc  can  be  placed 
on  the  triangle  ABO,  ca 
taking  the  direction  of 
OA  and  cb  of  OB.  The 
line  ab  will  then  divide 
the  sides  OA  and  OB  proportionally,  and  will  therefore 
be  parallel  to  AB,  and  the  triangles  will  be  equiangular 
and  similar,  (Th.  17). 

But  if  the  [_c  be  not  equal  to  the  [_  0,  then  place  ae 
on  i(J  as  before,  the  point  c  falling  on  0.     Under  the 
present  supposition  cb  will  not  fall  on  OB,  but  will  take 
another  direction,  OV,  on  one  side  or  the  other  of  OB 
Make  OV  equal  to  cb  and  draw  aV. 

Now,  the  A  abc  is  represented  in  magnitude  and  posi- 
tion by  the  A  a  VO;  and  if,  through  the  point  a,  the  line 
ab  be  drawn  parallel  to  AB,  we  shall  have 

Oa  :  OA  n  ab    :  AB; 
hut  by  (Hy.)        Oa  :  OA  : :  aV  :  AB. 


BOOK   II. 


79 


Hence,  (Th.  d), 

ab  :  AB  ::  aV  :  XB; 
which  requires  that  ab  =  aV,  but  (Th.  22,  B.  1)  ab  can 
not  be  equal  to  aV;  hence  the  last  proportion  is  absurd, 
and  the  supposition  that  the  [__  c  is  not  equal  to  the  [__  (7, 
which  leads  to  this  result,  is  also  absurd.  Therefore, 
the  [_c  is  equal  to  the  [__  (7,  and  the  triangles  are  equi- 
angular and  similar. 

Hence  the  theorem ;  if  any  triangle  have  its  sides,  etc. 


THEOREM    XIX. 

If  four  straight  lines  are  in  proportion,  the  rectangle  con- 
tained by  the  lines  which  constitute  the  extremes,  is  equivalent 
to  that  contained  by  those  which  constitute  the  means  of  the 
proportion. 

Let  A,  B,  C,  D,  represent  the  four      A'~ 
lines ; 


then 


T>   

we    are    to    show,   geo-      c, 
metrically,  that  A  x  JD  —  B  x  Q.  Dj 

Place  A  and  B  at  right  angles  to  each 
other,  and  draw  the  hypotenuse.  Also  place 
0  and  D  at  right  angles  to  each  other,  and 
draw  the  hypotenuse.  Then  bring  the  two 
triangles  together,  so  that  0  shall  be  at  right 
angles  to  B,  as  represented  in  the  figure. 

Now,  these  two  A's  have  each  a  E.  [_, 
and  the  sides  about  the  equal  angles  are  pro- 
portional ;  that  is,  A  :  B  : :  C  :  D ;  hence, 
(Th.  17,  Cor.  2),  the  two  A's  are  equiangular,  and  the 
acute  angles  which  meet  at  the  extremities  of  B  and  (7, 
are  together  equal  to  one  right  angle,  and  the  lines  B 
and  G  are  so  placed  as  to  make  another  right  angle; 
therefore,  also,  the  extremities  of  A,  B,  0,  and  D,  are  in 
one  right  line,  (Th.  3,  B.  I),  and  that  line  is  the  diag- 


l\    B 

\ 

BC 

\ 

C 

\» 

■   AD 

\ 

80  GEOMETRY. 

onal  of  the  parallelogram  be.  By  Th.  31,  B.  I,  the 
complementary  parallelograms  about  this  diagonal  are 
equal ;  but,  one  of  these  parallelograms  is  B  in  length, 
and  0  in  width,  and  the  other  is  D  in  length  and  A  in 
width;  therefore, 

B  x  C  =  A  x  B. 

Hence  the  theorem;  if  four  straight  lines  are  in  propor- 
tion, etc. 

Cor.  When  B  =  (7,  then  A  X  B  =  B\  and  B  is  the 
mean  proportional  between  A  and  D.  That  is,  if  three 
straight  lines  are  in  proportion,  the  rectangle  contained 
by  the  first  and  third  lines  is  equivalent  to  the  square 
described  on  the  second  line. 

THEOREM    XX. 

Similar  triangles  are  to  one  another  as  the  squares  of  their 

homologous  sides. 

Let  ABO  and  DBF  be  two 

similar  triangles,  and  LQ  and 

MF  perpendiculars  to  the  sides 

A B  and  BE  respectively.  Then 

we  are  to  prove  that 

aABC:aBEF  =  AB*:BE\ 
By  the  similarity  of  the  tri- 
angles, we  have, 

AB    :BE  =  LC  :  MF 
But,         AB   ;  BE  =  AB  :  BE 
Hence,     AB2  :  JJW  =  AB  x  LQ :  BE  x  MF. 
But,  (by  Th.  30,  B.  I),  AB  x  LQ  is  double  the  area 

of  the  A  ABC,  and  BE  x  MF  is  double  the  area  of  the 

A  BEF. 
Therefore,     A  ABO:  ABEFnAB  x  LQ  :BExMF 
And,  (Th.  6),  A  ABC:  A  BEF=  AW  :  BE\ 

Hence  the  theorem ;  similar  triangles  are  to  one  another, 

etc. 


BOOK    II. 


81 


The  following  illustration  will  enable  the  learner  fully 
to  comprehend  this  important  theorem,  and  it  will  also 
serve  to  impress  it  upon  his  memory. 

Let  abc  and  ABO  represent  two  equiangular  triangles. 
Suppose  the  length  of 
the  side  ac  to  be  two 
units,  and  the  length 
of  the  corresponding 
side  A  0  to  be  three 
units. 

Now,  drawing  lines 
through  the  points  of 

division  of  the  sides  ac  and  A  0,  parallel  to  the  other  sides 
of  the  triangles,  we  see  that  the  smaller  triangle  is  com- 
posed of  four  equal  triangles,  while  the  larger  contains 
nine  such  triangles.     That  is, 

the  sides  of  the  triangles  are  as  2  :  3, 

and  their  areas  are  as  4  :  9  =  22 :  3'. 


THEOREM   XXI 


Similar  polygons  may  be  divided  into  the  same  number  of 
triangles;  and  to  each  triangle  in  one  of  the  polygons  there 
will  be  a  corresponding  triangle  in  the  other  polygon,  these 
triangles  being  similar  and  similarly  situated. 

JjQtABCDUtmd.  abcde 
be  two  similar  polygons. 
Now  it  is  obvious  that  we 
can  divide  each  polygon 
into  as  many  triangles  as 
the  figure  has  sides,  less 

two;  and  as  the  polygons  have  the  same  number  of  sides, 
the  diagonals  drawn  from  the  vertices  of  the  homologous 
angles  will  divide  them  into  the  same  number  of  tri- 
angles. 


82  GEOMETRY. 

Since  the  polygons  are  similar,  the  angles  EAB  and  cab, 
are  equal,  and 

EA  :  AB  : :  ea  :  ab. 

Hence  the  two  triangles,  EAB  and  eab,  having  an  angle 
in  the  one  equal  to  an  angle  in  the  other,  and  the  sides 
about  these  angles  proportional,  are  equiangular  and 
similar,  and  the  angles  ABE  and  abe  are  equal. 

But  the  angles  ABO  and  abc  are  equal,  because  the 
polygons  are  similar. 

Hence,  [_ABO—  [_ABE=  \__abc  —  [__abe; 

that  is,  [___EBO  =  [__ebc. 

The  triangles,  EAB  and  eab,  being  similar,  their  ho- 
mologous sides  give  the  proportion, 

AB  :  BE  : :  ab  :  be;         ( 1 ) 
and  since  the  polygons  are  similar,  the  sides  about  the 
equal  angles  B  and  b  are  proportional,  and  we  have 
AB  :  BO  ::  ab  :  be; 

or,  BO  :  AB  ::  be  :  ab.       (2) 

Multiplying  proportions  (1)  and  (2),  term  by  term,  and 
omitting  in  the  result  the  factor  AB  common  to  the  terms 
of  the  first  couplet,  and  the  factor  ab  common  to  the 
terms  of  the  second,  we  have 

BO  :  BE  n  be  i  be. 
Hence  the  A's  EBO  and  ebe  are  equiangular  and  similar; 
and  thus  we  may  compare  all  of  the  triangles  of  one 
polygon  with  those  like  placed  in  the  other. 

Hence  the  theorem ;  similar  polygons  may  be  divided^  etc 

THEOREM    XXII. 

The  perimeter 8  of  similar  polygons  are  to  one  another  as 
their  homologous  sides  ;  and  their  areas  are  to  one  another  as 
the  squares  of  their  homologous  sides. 

Let  ABODE  and  abode  be  two  similar  polygons  ;  then 
we  are  to  prove  that  AB  is  to  the  sum  of  all  the  sides 


BOOK  II. 


of  the  polygon  A  BOB,  as 
ah  \%  to  the  sum  of  all 
the  sides  of  the  polygon 
abed.  E 

We  have  the  identical 
proportion 


AB  : 

ah  : : 

AB  : 

ah; 

and  since 

the 

polygons  are 

similar,  we  may  write  the 

following : 

AB  : 

ah  : 

1  BO  : 

he 

AB  : 

ah  : 

:  OB  : 

cd 

AB  : 

ah  : 

:  BE 

:  de9 

etc.  etc. 

Hence,  (Th.  7), 
AB  :  ah  ::  AB+BO+OD+DE,  etc.:  ab+bc+cd+de,  etc. 

Therefore,  the  perimeters  of  similar  polygons  are  to 
one  another  as  their  homologous  sides.  This  is  the  first 
part  of  the  theorem. 

Since  the  polygons  are  similar,  the  triangles  EAB,  eah, 
are  similar,  and  if  the  triangle  EAB  is  a  part  expressed 

by  the  traction  -,  of  the  polygon  to  which  it  belongs, 
n 

the  triangle  eah  is  a  like  part  of  the  other  polygon. 

Therefore,        EAB  :  eah  : :  ABOBEA  :  ahedea. 

But,  (Th.  20),  EAB  :  eah  : :  AB2  :  ah\ 

Therefore,  (Th.  6), 

ABOBEA  :  ahedea  : :  AB2  :  ah\ 

Therefore,  the  similar  polygons  are  to  one  another  as 
the  squares  on  their  homologous  sides.  This  is  the 
second  part  of  the  theorem. 

Hence  the  theorem  ;  the  perimeters  of  similar  polygon* 
are  to  one  another,  ete, 

THEOREM   XXIII. 

Two  triangles  which  have  an  angle  in  the  one  equal  tc  an 
angle  in  the  other,  are  to  each  other  as  the  rectangle  of  thu 
aides  about  the  equal  angles. 


84 


GEOMETRY. 


Let  ABC  and  def  be  two  triangles  having  the  angles 
A  and  d  equal.  It  is  to 
be  proved  that  the  areas 
ABC  and  def  are  to  each 
other  as  AB.AC  is  to 
de.df. 

Conceive  the  triangle 
def  placed  on  the  tri- 
angle ABC,  so  that  d 
shall  fall  on  A,  and  de  on 
AB ;  then  df  will  fall  on 
AC,  because  the  L'si 
and  d  are  equal.  On  AB,  lay  off  Ae,  equal  to  de ;  and 
on  AC,  lay  off  Af,  equal  to  df,  and  draw  ef  The  tri- 
angle Aef  will  then  be  equal  to  the  triangle  def  Join 
B  and/. 

Now,  as  triangles  having  the  same  altitude  are  to  each 
other  as  their  bases,  (Th.  16,  Cor.  1),  we  have 

Aef    :  ABf   ::  Ae  :  AB 
also,  ABf  :  ABC  : :  Af  :  AC 

Multiplying  these  proportions  together,  term  by  term, 
omitting  from  the  result  ABf,  a  factor  common  to  the 
terms  of  the  first  couplet,  we  have 

Aef  :  ABC  : :  Ae  .  Af  :  AB  .  AC 

But  Aef  is  equal  to  def,  Ae  to  de,  and  Af  to  df;  therefoie, 

def  :  ABC  ::  de  .  df  :  AB  .  AC 

Hence  the  theorem ;  two  triangles  which  have  an  angle,  ttc. 

Scholium.  —  If  we  suppose  that 

AB  :  AC  ::  de  :  df 

the  two  triangles  will  be  similar ;  and  if  we  multiply  the  terms  ot  the 
first  couplet  of  this  proportion  by  AC,  and  the  terms  of  the  second 
ccmplet  by  df,  we  shall  have 

AB  .  AC  :  AC%  : :  de ^  :  dj* 
U,  AB  .  AC  :  de  .  df  ::  AC2  :  df 


BOOK  II.  85 

Comparing  this  with  the  last  proportion  in  this  theorem,  and  we  have, 
(Th.6);  _       _ 

def:  ABC  xx  df  :  AC9 

Remark.  —  This  scholium  is  therefore  another  demonstration  of 
Theorem  20,  and  hence  that  theorem  need  not  necessarily  have  been 
made  a  distinct  proposition.  We  require  no  stronger  proof  of  the  cer- 
tainty of  geometrical  truth,  than  the  fact  that,  however  different  the 
processes  by  which  we  arrive  at  these  truths,  we  are  never  led  into 
inconsistencies ;  but  whenever  our  conclusions  can  be  compared,  they 
will  harmonize  with  each  other  completely,  provided  our  premises  are 
true  and  our  reasoning  logical. 

It  is  hoped  that  the  student  will  lose  no  opportunity  to  exercise 
his  powers,  and  test  his  skill  and  knowledge,  in  seeking  original 
demonstrations  of  theorems,  and  in  deducing  consequences  and 
conclusions  from  those  already  established. 

THEOREM   XXIV. 

If  the  vertical  angle  of  a  triangle  be  bisected,  the  bisecting 
line  will  cut  the  base  into  segments  proportional  to  the  adja- 
cent sides  of  the  triangle. 

Let  ABO  be  any  triangle, 
and  the  vertical  angle,  0,  be  bi-  *s1 

sected  by  tbe  straight  line  CD. 
Then  we  are  to  prove  that 
AB  :  BB  =  AC  :  OB. 

Produce  A O  to  E,  making    A  — D         r 

OE  =  OB,  and  draw  EB.  The  exterior  angle  A  OB,  of 
the  A  CEB,  is  equal  to  the  two  angles  E,  and  CBE; 
but  the  angle  E  =  QBE,  because  OB  —  OE,  and  the  tri- 
angle is  isosceles;  therefore  the  angle  AOB,  the  half  of 
the  angle  A  OB,  i3  equal  to  the  angle  E,  and  BO  and  BE 
are  parallel,  (Cor.2,Th.  7,B.  I). 

Now,  as  ABE  is  a  triangle,  and  OB  is  parallel  to  BE, 
we  have  AD  :  BB  =  AQ  :  OE  or  OB,  (Th.  17). 

Hence  the  theorem ;  if  the  vertical  angle  of  a  triangle 
be  bisected,  etc. 
8 


86  GEOMETRY. 

THEOREM   XXV. 

If  from  the  right  angle  of  a  right-angled  triangle,  a  per 
pendicular  is  drawn  to  the  hypotenuse ; 

1.  The  perpendicular  divides  the  triangle  into  two  similar 
triangles,  each  of  which  is  similar  to  the  whole  triangle, 

2.  The  perpendicular  is  a  mean  proportional  between  the 
segments  of  the  hypotenuse. 

3.  The  segments  of  the  hypotenuse  are  in  proportion  to  the 
squares  on  the  adjacent  sides  of  the  triangle. 

4.  The  sum  of  the  squares  on  the  two  sides  is  equivalent  to 
the  square  on  the  hypotenuse. 

Let  BAG  be  a  triangle,  right  an- 
gled at  A ;  and  draw  AD  perpendicu- 
lar to  BO. 

1.  The  two  A 's,  ABO  and  ABB,  B  DC 
have  the  common  angle,  B,  and  the  right  angle  BAQ  = 
the  right  angle  BDA ;  therefore,  the  third  |_  's  are  equal, 
and  the  two  A's  are  similar  by  Th.  17,  Cor.  1.  In  the 
same  manner  we  prove  the  A  ADC  similar  to  the  A 
ABC;  and  the  two  triangles,  ABB,  ABC,  being  similar 
to  the  same  A  ABO,  are  similar  to  each  other. 

2.  As  similar  triangles  have  the  sides  about  the  equal 
angles  proportional,  (Def.  16),  we  have 

BB  :  AD  ::  AD  :  OD; 

or,  the  perpendicular  is  a  mean  proportional  between  the  seg- 
ments of  the  hypotenuse. 

3.  Again,  BO j_BA  ::  BA  :  BD 
hence,  BA*  -  BO.BD        (1) 
also,  BQjJJA  : :  OA  :  OD 
hence,  OA2  =  BOOD         (2) 

Dividing  Eq.  (1)  by  Eq.  (2),  member  by  member,  wo 

obtain  

~BA*  ,     BD 

~OA2  ""   OD 


BOOK   II.  87 

which,  in  the  form  of  a  proportion,  is 

CA*  :~BA*  ::  CD  :  BD; 

that  is,  the  segments  of  the  hypotenuse  are  proportional  to  the 
squares  on  the  adjacent  sides. 
4.  By  the  addition  of  (1)  and  (2),  we  have 

BJl  +  CA*  m  BQ(BD  +  CD)  =BC\ 

that  is,  the  sum  of  the  squares  on  the  sides  about  the  right 
angle  is  equivalent  to  the  square  on  the  hypotenuse.  This  ia 
another  demonstration  of  Theorem  39,  B.  I. 

Hence  the  theorem ,  if  from  the  right  angle  of  a  right* 
angled  triangle,  etc. 


88 


GEOMETRY. 


BOOK  III. 


OF  THE  CIRCLE,  AND  THE  INVESTIGATION  OF  THEtt 
REMS  DEPENDENT  ON  ITS  PROPERTIES. 

DEFINITIONS. 


1.  *  A  Curved  Line  is  one  whose  consecutive  parts,  how- 
ever small,  do  not  lie  in  the  same  direction. 

2.  A  Circle  is  a  plane  figure  bounded  by  one  uniformly 
curved  line,  all  of  the  points  of  which  are  at  the  same 
distance  from  a  certain  point  within,  called  the  center 

3.  The  Circumference  of  a  cir- 
cle is  the  curved  line  that 
bounds  it. 

4.  The  Diameter  of  a  circle 
is  a  line  passing  through  the 
center,  and  terminating  at  both 
extremities  in  the  circumfer- 
ence. Thus,  in  the  figure,  0  is 
the  center  of  the  circle,  the 
curved  line  ACrBD  is  the  cir- 
cumference, and  AB  is  a  diameter. 

5.  The  Radius  of  a  circle  is  a  line  extending  from  the 
center  to  any  point  in  the  circumference.  Thus,  CD  is 
a  radius  of  the  circle. 

6.  An  Arc  of  a  circle  is  any  portion  of  the  circum- 
ference. 


*  The  first  six  of  the  above  definitions  have  been  before  given  among 
the  general  definitions  of  Geometry,  but  it  was  deemed  advisable  to 
reinsert  them  here. 


BOOK  III.  89 

7.  A  Chord  of  a  circle  is  the  line  connecting  the  ex- 
tremities of  an  arc. 

8.  A  Segment  of  a  circle  is  the  portion  of  the  circle  on 
either  side  of  a  chord. 

Tims,  in  the  last  figure,  EGrF  is  an  arc,  and  EF  is  a 
chord  of  the  circle,  and  the  spaces  bounded  by  the  chord 
EF,  and  the  two  arcs  EGrF  and  EDF,  into  which  it 
divides  the  circumference,  are  segments. 

9.  A  Tangent  to  a  circle  is  a  line  which,  meeting  the 
circumference  at  any  point,  will  not  cut  it  on  being 
produced.  The  point  in  which  the  tangent  meets  the 
circumference  is  called  the  point  of  tangency. 

10.  A  Secant  to  a  circle  is  a  line  which  meets  the  cir- 
cumference in  two  points,  and  lies  a  part  within  and  a 
part  without  the  circumference. 

11.  A  Sector  of  a  circle  is  a  portion  of  the  circle  included 
between  any  two  radii  and  their  intercepted  arc. 

Thus,  in  the  last  figure,  the  line  HL,  which  meets  the 
circumference  at  the  point  D,  but  does  not  cut  it,  is  a 
tangent,  D  being  the  point  of  tangency;  and  the  line 
MN,  which  meets  the  circumference  at  the  points  P  and 
Q,  and  lies  a  portion  within  and  a  portion  without  the 
circle,  is  a  secant.  The  area  bounded  by  the  arc  BD,  and 
the  two  radii  OB,  CD,  is  a  sector  of  the  circle. 

12.  A  Circumscribed  Polygon  is 
one  all  of  whose  sides  are  tangent 
to  the  circumference  of  the  circle ; 
and  conversely,  the  circle  is  then 
said  to  be  inscribed  in  the  polygon. 

13.  An  Inscribed  Polygon  is  one 
the  vertices  of  whose  angles  are 
all    found    in    the    circumference 
of  the  circle ;  and  conversely,  the  circle  is  then  said  to  be 
circumscribed  about  the  polygon. 

14.  A  Regular  Polygon  is  one  which  is  both  equiangu- 
lar and  equilateral. 

8* 


90 


GEOMETRY. 


The  last  three  definitions  are  illustrated  by  the  last 
figure. 

THEOREM    I. 

Any  radius  perpendicular  to  a  chord,  bisects  the  chord,  and 
also  the  arc  of  the  chord. 

Let  AB  be  a  chord,  0  the  center  of 
the  circle,  and  OE  &  radius  perpen- 
dicular to  AB ;  then  we  are  to  prove 
that  AB  =  BB,  and  AE  =  EB. 

Since  0  is  the  center  of  the  circle, 
AC—  BO,  CB  is  common  to  the  two 
A's  A  OB  and  BOB,  and  the  angles 
at  B  are  right  angles ;  therefore  the  two  A's  ABO  and 
BBO  are  equal,  and  AB  =  BB,  which  proves  the  first 
part  of  the  theorem. 

Now,  as  AB  —  BB,  and  BE  is  common  to  the  two 
spaces,  ABE  and  BBE,  and  the  angles  at  B  are  right 
angles,  if  we  conceive  the  sector  OBE  turned  over  and 
placed  on  CAE,  OE  retaining  its  position,  the  point  B 
will  fall  on  the  point  A,  because  AB  =  BB  and  AO  = 
BO;  then  the  arc  BE  will  fall  on  the  arc  AE;  otherwise 
there  would  be  points  in  one  or  the  other  arc  unequally 
distant  from  the  center,  which  is  impossible ;  therefore, 
the  arc  AE  —  the  arc  EB,  which  proves  the  second  part 
of  the  theorem. 

Hence  the  theorem. 

Cor.  The  center  of  the  circle,  the  middle  point  of 
the  chord  AB,  and  of  the  subtended  arc  AEB,  are 
three  points  in  the  same  straight  line  perpendicular  to 
the  chord  at  its  middle  point.  Now  as  but  one  perpen- 
dicular can  be  drawn  to  a  line  from  a  given  point  in  that 
line,  it  follows : 

1st.  That  the  radius  drawn  to  the  middle  point  3f 
any  arc  bisects,  and  is  perpendicular  to,  the  chord  of 
the  arc. 


BOOK    III 


91 


2d.  That  the  perpendicular  to  the  cl  ord  at  its  middle 
point  passes  through  the  center  of  the  circle  and  the 
middle  of  the  subtended  arc. 


THEOREM    II. 

Equal  angles  at  the  center  of  a  circle  are  subtended  by 
equal  chords. 

Let  the  angle  A  CE  =  the  angle 
ECB;  then  the  two  isosceles  triangles, 
ACE,  and  ECB,  are  equal  in  all  re- 
spects, and  AE  =  EB. 

Hence  the  theorem. 

THEOREM   III. 

In  the  same  circle,  or  in  equal  circles,  equal  chords  are 
equally  distant  from  the  center. 

Let  AB  and  EF  be  equal  chords, 
and  C  the  center  of  the  circle.  From 
C,  draw  CG  and  CH,  perpendicular 
to  the  respective  chords.  These 
perpendiculars  will  bisect  the  chords, 
(Th.  1),  and  we  shall  have  A  G  =  EH. 
"We  are  now  to  prove  that  CG  =  CH. 

Since  the  A's  ECH  and  AQG  are  right-angled,  we 
have,  (Th.  39,  B.  I), 

EH2  +~WC7  =-EC2 

and,  AG2  +  GO2  =  'AC2. 

By  subtracting  these  equations,  member  from  mem- 
ber, we  find  that 

EH2  — AG2  +  7W2 —  ~GC2  =~EC2 —  AC*    (1) 
But  the  chords  are  equal  by  hypothesis,  hence  their 
halves,  EH  and  AG,  are  equal;  also  EC  =  AC,  being 
radii  of  the  circle.     Wherefore, 


92  GEOMETRY. 

EH2  —  AG2  «  0 
and,  W  —  ~A02  -  0. 

These  values  in  Equation  ( 1 )  reduce  it  to 

HC?  -  GO2  =  0 
or,  H02=G02 

and,  #(7  -  #(7. 

Hence  the  theorem. 
(7<?r.    Under  all  circumstances  we  have 
EH2  +  HQ2  =  Z#2  +  #tf 2, 

because  the  sum  of  the  squares  in  either  member  of  the 
equation  is  equivalent  to  the  square  of  the  radius  of  the 
circle. 

Now,  if  we  suppose  HO  greater  than  GO,  then  will 
HQ2  be  greater  than  GO2.  Let  the  difference  of  these 
squares  be  represented  by  d. 

Subtracting  GO2  from  both  members  of  the  above 
equation,  we  have 

EH2+d=AG2 

whence,        ~AG2 >  EH2,  and  A G > EH. 

Therefore,  AB,  the  double  of  AG,  is  greater  than  EF, 
the  double  of  EH;  that  is,  of  two  chords  in  the  same  or 
equal  circles,  the  one  nearer  the  center  is  the  greater. 

The  equation,  EH2  +  HO2  =  AG2  -f  ~G02,  being  true, 
whatever  be  the  position  of  the  chords,  we  may  suppose 
GO  to  have  any  value  between  0  and  A 0,  the  radius  of 
the  circle. 

When  GO  becomes  zero,  the  equation  reduces  to 
EH2  +  ~H02  -  AG*  -  B2; 
that  is,  under  this  supposition,  AG  coincides  with  A 0, 
and  AB  becomes  the  diameter  of  the  circle,  the  greatest 
chord  that  can  he  drawn  in  it. 


BOOK    111.  93 

THEOREM   IV 

A  line  tangent  to  the  circumference  of  a  circle  is  at  right 
angles  with  the  radius  drawn  to  the  point  of  contact. 

Let  A  0  be  a  line  tangent  to  the  circle 
at  the  point  B,  and  draw  the  radius,  EB, 
and  the  lines,  AE  and  OE. 

Now,  we  are  to  prove  that  EB  is  per- 
pendicular to  AC.  Because  B  is  the 
only  point  in  the  line  A 0  which  meets 
the  circle,  (Def.  9,  B.  Ill),  any  other  line, 
as  AE  or  CE,  must  be  greater  than  EB ; 
therefore,  EB  is  the  shortest  line  that  can  be  drawn  from 
the  point  E  to  the  line  A  0;  and  EB  is  the  perpendicu- 
lar to  AC,  (Th.  23,  B.  I). 

Hence  the  theorem. 

THEOREM   V. 

In  the  same  circle,  or  in  equal  circles,  equal  chords  subtend 
or  stand  on  equal  portions  of  the  circumference. 

Conceive  two  equal  circles,  and  two  equal  chords  drawn 
within  them.  Then,  conceive  one  circle  taken  up  and 
placed  upon  the  other,  center  upon  center,  in  such  a  po- 
sition that  the  two  equal  chords  will  fall  on,  and  exactly 
coincide  with,  each  other;  the  circles  must  also  coin- 
ciie,  because  they  are  equal;  and  the  two  arcs  of  the  two 
circles  on  either  side  of  the  equal  chords  must  also  coin- 
cide, or  the  circles  could  not  coincide;  and  magnitudes 
which  coincide,  or  exactly  fill  the  same  space,  are  in  all 
respects  equal,  (Ax.  10). 

Hence  the  theorem. 


94  GEOMETRY. 

THEOREM   VI. 

Through  three  given  points,  not  in  the  same  straight  line, 
one  circumference  can  be  made  to  pass,  and  but  one. 

Let  A,  B,  and  0  be  three  given 

points,  not  in  the  same  straight 

line,  and  draw  the  lines  AB  and 

BO.    If  a  circumference  is  made 

to  pass  through  the  two  points  A 

and  B,  the  line  AB  will  be  a  chord 

to  such  a  circle ;  and  if  a  chord  is 

bisected  by  a  line  at  right  angles, 

the  bisecting  line  will  pass  through 

the  center  of  the  circle,  (Cor.,  Th.  1) ;  therefore,  if  we 

bisect  the  line  AB,  and  draw  DF,  perpendicular  to  AB, 
at  the  point  of  bisection,  any  circumference  that  can 
pass  through  the  points,  A  and  B,  must  have  its  center 
somewhere  in  the  line  DF.  And  if  we  draw  FIG-  at 
right  angles  to  BO  at  its  middle  point,  any  circumference 
that  can  pass  through  the  points  B  and  0  must  have  its 
center  somewhere  in  the  line  EO.  Now,  if  the  two  lines, 
DF  and  EGr,  meet  in  a  common  point,  that  point  will  be 
a  center,  about  which  a  circumference  can  be  drawn  to 
pass  through  the  three  points,  A,  B,  and  0,  and  DF  and 
EO  will  meet  in  every  case,  unless  they  are  parallel ;  but 
they  are  not  parallel,  for  if  they  were,  it  would  follow 
(Th.  5,  B.  I)  that,  since  DF  is  intersected  at  right  angles 
by  the  line  AB,  it  must  also  be  intersected  at  right  angles 
by  the  line  BO,  having  a  direction  different  from  that  of 
AB ;  which  is  impossible,  (Th.  7,  B.  I). 

Therefore  the  two  lines  will  meet ;  and,  with  the  point 
H,  at  which  they  meet,  as  a  center,  and  HB  —  HA  =  HQ 
as  a  radius,  one  circumference,  and  but  one,  can  be  made 
to  pass  through  the  three  given  points. 
Hence  the  theorem. 


BOCK   III.  95 

THEOREM  VII. 

If  two  rircles  touch  each  other,  either  internally  or  exter- 
nally, the  two  centers  and  the  point  of  contact  will  be  in  one 
right  line. 

Let  two  circles  touch  each 
other  internally,  as  represented 
at  A,  and  conceive  AB  to  be  a 
tangent  at  the  common  point  A. 
Now,  if  a  line,  perpendicular  to 
AB,  be  drawn  from  the  point 
A,  it  must  pass  through  the 
center  of  each  circle,  (Th.  4) ; 

and  as  but  one  perpendicular  can  be  drawn  to  a  line  at  a 
given  point  in  it,  A,  0,  and  B,  the  point  of  contact  and 
the  two  centers  must  be  in  one  and  the  same  line. 

Next,  let  two  circles  touch  each  other  externally,  and 
from  the  point  of  contact  conceive  the  common  tangent, 
AB,  to  be  drawn. 

Then  a  line,  A  0,  perpendicular  to  AB,  will  pass 
through  the  center  of  one  circle,  (Th.  4),  and  a  per- 
pendicular, AB,  from  the  same  point,  A,  will  pass 
through  the  center  of  the  other  circle ;  hence,  BAG  and 
BAB  are  together  equal  to  two  right  angles ;  therefore 
CAB  is  one  continued  straight  line,  (Th.  3,  B.  I). 

Cor.  "When  two  circles  touch  each  other  internally,  the 
distance  between  their  centers  is  equal  to  the  difference 
of  their  radii ;  and  when  they  touch  each  other  extern- 
ally, the  distance  between  their  centers  is  equal  to  the 
sum  of  their  radii. 

THEOREM   VIII. 

An  angle  at  the  circumference  of  any  circle  is  measured  by 
one  half  the  arj  on  which  it  stands. 

In  this  work  it  is  taken  as  an  axiom  that  any  angle 
whose  vertex  is  at  the  center  of  a  circle,  is  measured  by 


96 


GEOMETRY. 


the  arc  on  which  it  stands ;  and  we  now  proceed  to  prove 
that  when  the  arcs  are  equal,  the  angle  at  the  circumference 
is  equal  to  one  half  the  angle  at  the  center. 

Let  ACB  be  an  angle  at  the  center, 
and  D  an  angle  at  the  circumference, 
and  at  first  suppose  D  in  a  line  with 
AC.  We  are  now  to  prove  that  the 
angle  ACB  is  double  the  angle  D. 

The  A  DCB  is  an  isosceles  triangle, 
because  CD  =  CB ;  and  its  exterior 
angle,  ACB,  is  equal  to  the  two  interior  angles,  D,  and 
CBD,  (Th.  12,  B.  I),  and  since  these  two  angles  are  equal 
to  each  other,  the  angle  ACB  is  double  the  angle  at 
D.  But  ACB  is  measured  by  the  arc  AB ;  therefore  the 
angle  D  is  measured  by  one  half  the  arc  AB. 

Next,  suppose  D  not  in  a  line  with 
AC,  but  at  any  point  in  the  circum- 
ference, except  on  AB ;  produce  DC 
to  E. 

Now,  by  the  first  part  of  this 
theorem, 

the  angle  ECB  =  2EDB, 

also,  EC  A  =  2EDA, 

by  subtraction,  ACB  =  2  ADB. 

But  ACB  is  measured  by  the  arc  AB;  therefore  ADB 
or  the  angle  D,  is  measured  by  one  half  of  the  same  arc 
Hence  the  theorem. 


THEOREM   IX. 

An  angle  in  a  semicircle  is  a  right  angle ;  an  angle  in  a 
(segment  greater  than  a  semicircle  is  less  than  a  right  angle  ; 
and  an  angle  in  a  segment  less  than  a  semicircle  is  greater 
than  a  right  angle. 

If  the  angle  ACB  is  in  a  semicircle,  the  opposite  seg- 
ment, ADB,  on  which  it  stands,  is  also  a  semicircle ;  and 
the  angle  ACB  is  measured  by  one  half  the  arc  ADB 


BOOK    III. 


b« 


(Th.  8) ;  that  is,  one  half  of  180°,  or  90°,  which  is  the 
measure  of  a  right  angle. 

If  the  angle  A  CB  is  in  a  segment 
greater  than  a  semicircle,  then  the 
opposite  segment  is  less  than  a  semi- 
circle, and  the  measure  of  the  angle 
is  less  than  one  half  of  180°,  or  less 
than  a  right   angle.     If  the   angle 
ACB  is  in  a  segment  less  than  a 
semicircle,  then  the  opposite  segment,  ABB,  on  whiefc 
the  angle  stands,  is  greater  than  a  semicircle,  and  its  half 
is  greater  than  90°;   and,  consequently,  the  angle  is 
greater  than  a  right  angle. 

Hence  the  theorem. 

Cor.  Angles  at  the  circumference, 
and  standing  on  the  same  arc  of  a 
circle,  are  equal  to  one  another ;  for 
all  angles,  as  BAG,  BBC,  BBC,  are 
equal,  because  each  is  measured  by 
one  half  of  the  arc  BC.  Also,  if  the 
angle  BEC  is  equal  to  CEG-,  then 
the  arcs  Bd  and  CG-  are  equal,  be- 
cause their  halves  are  the  measures  of  equal  angles. 

THEOREM    X. 

The  sum  of  two  opposite  angles  of  any  quadrilateral  in* 
scribed  in  a  circle,  is  equal  to  two  right  angles. 

Let  ACBD  represent  any  quadri- 
lateral inscribed  in  a  circle.  The 
angle  ACB  has  for  its  measure,  one 
half  of  the  arc  ABB,  and  the  angle 
ABB  has  for  its  measure,  one  half  of 
the  arc  ACB;  therefore,  by  addition, 
the  sum  of  the  two  opposite  angles  at 
C  and  B,  are  together  measured  by 
one  half  of  the  whole  circumference,  or  by  180  degrees, 
«=  two  right  angles.     Hence  the  theorem 


98 


GEOMETRY. 


THEOREM  XI. 

An  angle  formed  by  a  tangent  and  a  chord  is  measured  ty 
one  half  of  the  intercepted  arc. 

Let  AB  be  a  tangent,  and  AD  a 
chord,  and  A  the  point  of  contact ; 
then  we  are  to  prove  that  the  angle 
BAD  is  measured  by  one  half  of  the 
arc  AED. 

From  A  draw  the  radius  AC;  and 
from  the  center,  C,  draw  CE  per- 
pendicular to  AD. 

The    [_BAD  +  [__DAC  =  90°,  (Th.  4). 

Also,        [_C+l_DAC=  90°,  (Cor.  4,  Th.  12,  B.  1), 

Therefore,  by  subtraction,  BAD  —  (7=0; 

by  transposition,  the  angle         BAD  =  C. 

But  the  angle  C,  at  the  center  of  the  circle,  is  measured 
by  the  arc  AE,  the  half  of  AED ;  therefore,  the  equal 
angle,  BAD,  is  also  measured  by  the  arc  AE,  the  half 
of  AED. 

Hence  the  theorem. 


THEOREM   XII. 

An  angle  formed  by  a  tangent  and  a  chord,  is  equal  to  an 
angle  in  the  opposite  segment  of  the  circle. 

Let  AB  be  a  tangent,  and  AD  a 
chord,  and  from  the  point  of  contact, 
A,  draw  any  angles,  as  AOD,  and 
AED,  in  the  segments.  Then  we  are 
to  prove  that  [__  BAD  =  [__  A  CD,  and 
L  GAD  =  L  AED- 

By  Th.  11,  the  angle  BAD  is  meas- 
ured by  one  half  the  arc  AED ;  and 
as  the  angle  A  CD  is  measured  by  one  half  of  the  same 
arc,  (Th.  8),  we  have  |_  BAD  =  [_  ACD. 


BOOK   III.  90 

Again,  as  AEBO  is  a  quadrilateral,  inscribed  in  a 
circle,  the  sum  of  the  opposite  angles, 

AOB  +  AEB  -  2  right  angles.     (Th.  10). 

Also,  the  sum  of  the  angles 

BAD  +  BAG  =  2  right  angles.     (Th.  1,  B.  I). 

By  subtraction  (and  observing  that  BAB  has  just  been 
proved  equal  to  AOB),  we  have, 

AEB  — BAG  =  0. 
Or,  by  transposition,         AEB  =  BAG* 

Hence  the  theorem. 

THEOREM    XIII. 

Arcs  of  the  circumference  of  a  circle  intercepted  by  paral- 
lel chords,  or  by  a  tangent  and  a  parallel  chord,  are  equal. 

Let  AB  and  OB  be  parallel  chords, 
and  draw  the  diagonal,  AB ;  now,  be- 
cause AB  and  CB  are  parallel,  the 
angle  BAB  -  the  angle  ABO  (Th.  6,  B. 
I) ;  but  the  angle  BAB  has  for  its  meas- 
ure, one  half  of  the  arc  BB\  and  the 
angle  ABO  has  for  its  measure,  one  half  of  the  arc  AO, 
(Th.  8) ;  and  because  the  angles  are  equal,  the  arcs  are 
equal ;  that  is,  the  arc  BB  =  the  arc  A  0. 

Next,  let  EF  be  a  tangent,  parallel  to  a  chord,  OB,  and 
from  the  point  of  contact,  G,  draw  GB. 

Since  EF  and  OB  are  parallel,  the  angle  OBG  =  the 
angle  BGF.  But  the  angle  OBGr  has  for  its  measure, 
one-half  of  the  arc  00-,  (Th.  8) ;  and  the  angle  BGF 
has  for  its  measure,  one  half  of  the  arc  GB,  (Th.  11) ; 
therefore,  these  measures  of  equals  must  be  equal ;  that 
is,  the  arc  CG=the  arc  GD. 

Hence,  the  theorem. 


100  GEOMETRY. 


THEOREM    XIV. 


When  two  chords  intersect  each  other  within  a  circle,  the 
angle  thus  formed  is  measured  by  one  half  the  sum  of  the  twa 
intercepted  arcs. 

Let  AB  and  CD  intersect  each 
other  within  the  circle,  forming  the 
two  angles,  E  and  Ef,  with  their 
equal  vertical  angles. 

Then,  we   are  to  prove  that  the 
angle  E  is  measured  by  one  half  the 
sum  of  the  arcs  AC  and  BD;  and 
the  angle  E'  is  measured  by  one  half  the  sum  of  the 
arcs  AD  and  CB. 

First,  draw  AF  parallel  to  CD,  and  FD  will  be  equal 
to  AC,  (Th.  13);  then,  by  reason  of  the  parallels,  [__  BAF 
=  |__  E.  But  the  angle  BAF  is  measured  by  one  half 
of  the  arc  BDF;  that  is,  one  half  of  the  arc  BD  plus  one 
half  of  the  arc  AC. 

Now,  as  the  sum  of  the  angles  E  and  Ef  is  equal  to 
two  right  angles,  that  sum  is  measured  by  one  half  the 
whole  circumference. 

But  the  angle  E,  alone,  as  we  have  just  proved,  is 
measured  by  one  half  the  sum  of  the  arcs  BD  and  AC; 
therefore,  the  other  angle,  E',  is  measured  by  one  half 
the  sum  of  the  other  parts  of  the  circumference, 
AD  +  CB. 

Hence  the  theorem. 

THEOREM   XV. 

When  two  secants  intersect,  or  meet  each  other  without  a 
circle,  the  angle  thus  formed  is  measured  by  one  half  the  dif 
ference  of  the  intercepted  arcs.  ^ 


BOOK    III. 


101 


Let  DE  and  BE  be  two  secants 
meeting  at  E ;  and  draw  A F  parallel  to 
CD.  Then,  by  reason  of  the  parallels, 
the  angle  E,  made  by  the  intersection 
of  the  two  secants,  is  equal  to  the 
angle  BAF.  But  the  angle  BAF  is 
measured  by  one  half  the  arc  BF; 
that  is,  by  one  half  the  difference  be- 
tween the  arcs  BD  and  A  C. 

Hence  the  theorem. 


THEOREM   XVI. 

The  angle  formed  by  a  secant  and  a  tangent  is  measured 
by  one  half  the  difference  of  the  intercepted  arcs. 

Let  BQ  be  a  secant,  and  CD  a  tan- 
gent, meeting  at  C.  We  are  to  prove 
that  the  angle  formed  at  C,  is  meas- 
ured by  one  half  the  difference  of  the 
arcs  BD  and  DA. 

From  A,  draw  AE  parallel  to  CD ; 
then  the  arc  AD  =  the  arc  DE; 
BD  --DE  =  BE;  and  the  [__ BAE  = 
L  O.  But  the  angle  BAE  is  measured 
by  one  half  the  arc  BE,  (Th.  8,)  that  is,  by  one  half 
the  difference  between  the  arcs  BD  and  AD;  there- 
fore, the  equal  angle,  C,  is  measured  by  one  half  the 
arc  BE. 

Hence  the  theorem. 


THEOREM    XVII. 

When  two  chords  intersect  each  other  in  a  circle,  the  rect- 
angle contained  by  the  segments  of  the  one,  will  be  equivahrt 
to  the  rectangle  contained  by  the  segments  of  the  other. 
9* 


102 


GEOMETRY. 


Let  AB  and  CB  be  two  chords  inter- 
Becting  each  other  in  E.  Then  we  are 
to  prove  that  the  rectangle  AE  x  EB  = 
the  rectangle  CE  X  EB. 

Draw  the  lines  AB  and.  CB,  forming 
the  two  triangles  AEB  and  CEB.  The 
angles  B  and  B  are  equal,  because  they 
are  each  measured  by  one  half  the  arc,  AG.  Also  the 
angles  A  and  C  are  equal,  because  each  is  measured  by 
one  half  the  arc,  BB ;  and  L  AEB  =  |__  CEB,  because 
they  are  vertical  angles ;  hence,  the  triangles,  AEB  and 
CEB,  are  equiangular  and  similar.  But  equiangular  tri- 
angles have  their  sides  about  the  equal  angles  propor- 
tional, (Cor.  1,  Th.  17,  B.  II);  therefore,  AE  and  EB, 
about  the  angle  E,  are  proportional  to  CE  and  EB,  about 
the  same  or  equal  angle. 

That  is,  AE  :  EB  ::  CE  :  EB; 

Or,  (Th.  19,  B.  II),      AE  x  EB  =  CE  x  EB. 

Hence  the  theorem. 

Cor.  When  one  chord  is  a  diameter,  and  the  other  at  right 
angles  to  it,  the  rectangle  contained  by  the  segments  of  the 
diameter  is  equal  to  the  square  of  one  half  the  other  chord ; 
or  one  half  of  the  bisected  chord  is  a  mean  proportional  be- 
tween the  segments  of  the  diameter. 

For,  ABxBB*=FBx  BE.  But,  if 
AB  passes  through  the  center,  C,  at 
right  angles  to  FE,  then  FB  =  BE 
(Th.  1) ;  and  in  the  place  of  FB,  write 
its  equal,  BE,  in  the  last  equation,  and 
we  have 

ABxBB  =  EE\ 

or,  (Th.  3,  B  II),    AB  :  BE  : :  BE  :  BB. 

Put,  BE  =  x,  CB  ==  y,  and  CE  =  B,  the  radius  of  the 
circle. 


BOOK    III. 


103 


Then  AD  ~  B  -  -y,  and  DB  -  B  -f  #.  With  this  nota- 
tion, 

AD  x  DB  =  DE% 

becomes,  (B  -y)(R  +  y)  =  x* 

or,  B%  —  y2  =  x* 

or,  B2  =  x*+y* 

That  is,  f  A«  square  of  the  hypotenuse  of  the  right-angled 
triangle,  DCE,  is  equal  to  the  sum  of  the  squares  of  the  other 
two  sic 


THEOREM    XVIII. 

If  from  a  point  without  a  cirde,  a  tangent  line  be  drawn  to 
the  circumference,  and  also  any  secant  line  terminating  in  tht- 
concave  arc,  the  square  of  the  tangent  will  be  equivalent  to  b*. 
rectangle  contained  by  the  whole  secant  and  its  external  sea 
ment. 

Let  A  be  a  point  without  the 
circle  DUG,  and  let  AD  be  a 
tangent  and  AE  any  secant  line. 

Then  we  are  to  prove  that 
AOxAE  =  AD\ 
In  the  two  triangles,  ADE  and 
ADC,  the  angles  ADO  and  AED 
are  equal,  since  each  is  meas- 
ured by  one  half  of  the  same 
arc,  DO;  the  angle  A  is  com- 
mon to  the  two  triangles ;  their 

third  angles  are  therefore  equal,  and  the  triangles  are 
equiangular  and  similar. 

Their  homologous  sides  give  the  proportion 
AE  :  AD  i  :_AZ)  :  AC 

whence,  AE  xAC  =  AD2 

Hence  the  theorem. 

Cor.  If  AE  and  AF  are  two  secant  lines  drawn  from 
%\  e  same  point  without  the  circumference,  we  shall  have 


104  GEOMETRY. 

AOx  AF=AB* 
and,  ABxAF^AB* 

hence,  AOx  AF  =  AB  x  AF, 

which,  in  the  form  of  a  proportion,  gives 
AC  :  AF  ::AB  :  AF. 
That  is,  the  secants  are  reciprocally  proportional  to  their  ex- 
ternal segments. 

Scholium.  —  By  means  of  this  theorem  we  can  determine  the  diam- 
eter of  a  circle,  when  we  know  the  length  of  a  tangent  drawn  from  a 
point  without,  and  the  external  segment  of  the  secant,  which,  drawn 
from  the  same  point,  pasees  through  the  center  of  the  circle. 

Let  Am  be  a  secant  passing  through  the  center,  and 
suppose  the  tangent  AB  to  he  20,  and  the  external  seg- 
ment, An,  of  the  secant  to  be  2.  Then,  if  D  denote  the 
diameter,  we  shall  have 

im=2+i), 

whence,    Am  x  An  =  2  (2  +  B)  =  4  -f  2B  =  (20)2  =  400, 
22)  =  396,  and  B  =  198. 

Ifiw,  the  height  of  a  mountain  on  the  earth,  and  AB, 
the  distance  of  the  visible  sea  horizon,  be  given,  we  may 
determine  the  diameter  of  the  earth. 

For  example ;  the  perpendicular  height  of  a  mountain 
on  the  island  of  Teneriffe  is  about  3  miles,  and  its  summit 
can  be  seen  from  ships  when  they  are  known  to  be  154 
or  155  miles  distant ;  what  then  is  the  diameter  of  the 
earth  ? 

Designate,  as  before,  the  diameter  by  B.  Then  A  m  «■ 
3  +  2),  and  Am  x  An  =  9  +  SB.  AB  =  154.  5 ;  hence, 
9  +  32)  =  (154.  5)2  =  23870.  25,  from  which  we  find  2)  = 
7953.75,  which  differs  but  little  from  the  true  diameter* 
of  the  earth. 

One  source  of  error,  in  this  mode  of  computing  the 
diameter  of  the  earth,  is  atmospheric  refraction,  the  ex 
planation  of  which  does  not  belong  here. 


HOOK   III.  105 

THEOREM    XIX. 

If  a  circle  be  described  about  a  triangle,  the  rectangle  con- 
tained by  two  sides  of  the  triangle  is  equivalent  to  the  rectangle 
contained  by  the  perpendicular  let  fall  on  the  third  side,  and 
the  diameter  of  the  circumscribing  circle. 

Let  ABO  be  a  triangle,  AC  and 
OB,  the  side3,  OB  the  perpendicular 
let  fall  on  the  base  AB,  and  OB  the 
diameter  of  the  circumscribing  circle. 
Then  we  are  to  prove  that 

AO  x  OB=  OEx  OB. 

The  two  A's,  A  OB  and  OEB,  are 
equiangular,  because  \__A=\__E,  both 
being  measured  by  the  half  of  the  arc  OB;  also,  ABO  is 
a  right  angle,  and  is  equal  to  OBB,  an  angle  in  a  semi- 
circle, and  therefore  a  right  angle ;  hence,  the  third  angle, 
AOB  =  [_BOE,  (Th.  12,  Cor.  2,  B.  I).  Therefore,  (Cor.  1, 
Th.  17,  B.  II), 

AO  x  OB  ::  OE  :  OB 

and,  AOxBO=OEx  OB. 

Hence  the  theorem ;  if  a  circle,  etc. 

Oor.  The  continued  product  of  three  sides  of  a  triangle  is 
equal  to  twice  the  area  of  the  triangle  into  the  diameter  of  its 
tircumscribing  circle. 

Multiplying  both  members  of  the  last  equation  by  A  B, 
we  have, 

AOx  BOxAB=OEx  {AB  x  OB). 

But  OB  is  the  diameter  of  the  circle,  and  (AB  x  OB) 
«=  twice  the  area  of  the  triangle ; 

Therefore,  AOx  OB  x  AB=  diameter  multiplied 
by  twice  the  area  of  the  triangle. 


106  GEOMETRY. 

THEOREM   XX. 

The  square  of  a  line  bisecting  any  angle  of  a  triangle,  to* 
gether  with  the  rectangle  of  the  segments  into  which  it  cuts  the 
opposite  side,  is  equivalent  to  the  rectangle  of  the  two  sides 
including  the  bisected  angle. 

Let  ABO  be  a  triangle,  and  CD  a 
Line  bisecting  the  angle  C.  Then 
we  are  to  prove  that 

CD*  +  (AD  x  DB)  =  ACx  CB. 
The  two  A's,  ACE  and  CDB,  are 
equiangular,  because  the  angles  E 
and  B  are  equal,  both  being  in  the 
same  segment,  and  the  [__  ACE  =  BCD,  by  hypothesis. 
Therefore,  (Th.  17,  Cor.  1,  B.  H), 

AC  :  CE  ::  CD  :  CB. 
But  it  is  obvious  that  CE =  CD  -f  DE,  and  by  substi- 
tuting this  value  of  CE,  in  the  proportion,  we  have, 
AC  i  CD  +  DE  ::  CD  :  CB. 
By  multiplying  extremes  and  means, 

CD2  +  (DE  x  CD)  =  ACx  CB. 
But  by  (Th.  17), 

DE  x  CD  =  ADx  DB, 
and  substituting,  we  have, 

HB*  +  (AD  x  DB)  =  ACx  CB. 
Hence  the  theorem. 

THEOREM   XXI. 

The  rectangle  contained  by  the  two  diagonals  of  any  quad- 
rilateral inscribed  in  a  circle,  is  equivalent  to  the  sum  of  the 
two  rectangles  contained  by  the  opposite  sides  of  the  quadri- 
lateral. 

Let  ABCD  be  a  quadrilateral  inscribed  in.  a  circle; 
then  we  are  to  prove  that 

ACx  BD  =  (AB  x  DC)  +  (AD  x  BC). 

From  0,  draw  CE,  making  the  angle  DOE  equal  to 


BOOK    III.  107 

the  angle  A  OB ;  and  as  the  angle  BAO  is  equal  to  th* 
angle  ODE,  both  being  in  the  same  seg- 
ment, therefore,  the  two  triangles,  BEO 
and  ABO,  are  equiangular,  and  we  have 
(Th.  IT,  Cor.  1,  B.  II), 

AB  :  AC  ::  BE  :  BO    (1) 
The  two  A's,  ABO  and  BEO,  are 
equiangular;  for  the  [_BAO=  \_EBO, 
both  being  in  the  same  segment;  and  the  L  BOA  — 
\_EOB,  for  BOE  =  BOA;  to  each  of  these  add  the  ai  gle 
EOA,  smdL  BOA  =  EOB;   therefore,   (Th.  17,    Cor.  1, 

B.II), 

AB  :  AO  ::  BE  :  BO    (2). 

By  multiplying  the  extremes  and  means  in  proportions 
(1)  and  (2),  and  adding  the  resulting  equations,  we  have, 

(AB  x  BO)  +  (AB  x  BO)  =  (BE  +  BE)  x  AO. 
But,  BE  +  BE  =  BB;  therefore, 

(AB  x  2)(7)  -f  (AB  x  5(7)  =  i(Jx  BB, 

Oor.  When  two  adjacent  sides  of  the  quadrilateral  are 
equal,  as  AB  and  BO,  then  the  resulting  equation  is, 

(AB  x  BO)  +  (AB  x  AB)  =  AO  x  5D; 

or,  -15  x  (2)<7  -f  AB)  =  AO  x  BB; 

or,  AB  :  AO  : :  BB  :  BO  +  AB. 

That  is,  one  of  the  two  equal  sides  of  the  quadrilateral 
is  to  the  adjoining  diagonal,  as  the  transverse  diagonal  is  to 
the  sum  of  the  two  unequal  sides. 

THEOREM    XXII. 

If  two  chords  intersect  each  other  at  right  angles  in  a  cir- 
cle, the  sum  of  the  squares  of  the  four  segments  thus  formed 
is  equivalent  to  the  square  of  the  diameter  of  the  circle. 

Let  AB  and   OB  be   two   chords,  intersecting  eacb 
other  at  right  angles.     Draw  BE  parallel  to  EB,  and 
draw  BE  and  AF.    Now,  we  are  to  prove  that 
IE1  +^F  +~W%  f  EB2  =~AF\ 


108 


GEOMETRY. 


As  BF  is  parallel  to  ED,  ABF  is  a 
right  angle,  and  therefore  AF  is  a  diam- 
eter, (Th.  9).  Also,  because  BF  is 
parallel  to  CD,  OB  =  DF,  (Th.  13). 

Because  CEB  is  a  right  angle, 

CE2  +  EB2=CB2  =  DF\ 

Because  AED  is  a  right  angle, 

•    AE2 +~ED2  =  AD\ 

Adding  these  two  equations,  we  have, 

W2  +  EB2  +  AE2  +~ED2  =  DF2  +  ~AD\ 

But,  as  AF  is  a  diameter,  and  ADF  a  right  ang\e, 
(Th.  9),  _       _ 

DT+AD2  =  AF2; 

therefore,     CE2  +  EB2  -f  AE2  +  Ji)2  =  ZF2. 
Hence  the  theorem. 

Scholium.  —  If  two  chords  intersect  each  other  at  right  angles,  in  a 
circle,  and  their  opposite  extremities  be  joined,  the  two  chords  thus 
formed  may  make  two  sides  of  a  right-angled  triangle,  of  which  tho 
diameter  of  the  circle  is  the  hypotenuse. 

For,  AD  is  one  of  these  chords,  and  CB  is  the  other ;  and  we  have 
shown  that  CB  =  DF;  and  AD  and  DF  are  two  sides  of  a  right- 
angled  triangle,  of  which  AF  is  the  hypotenuse ;  therefore,  AD  and 
CB  may  be  considered  the  two  sides  of  .a  right-angled  triangle,  and 
AF  its  hypotenuse. 

THEOREM  XXIII. 

If  two  secants  intersect  each  other  at  right  angles,  the  sum 
of  their  squares,  increased  by  the  sum  of  the  squares  of  the 
two  segments  without  the  circle,  will  be  equivalent  to  the  square 
of  the  diameter  of  the  circle. 

Let  AE  and  ED  be  two  secants  in- 
tersecting at  right  angles  at  the  point 
E.  From  B,  draw  BF  parallel  to  CD, 
and  draw  AF  and  AD.  Now  we  are  to 
prove  that 

f3M   ED2  -r  EB2  +~Wf  T  IF. 


BOOK  III.  109 

Because  BF  is  parallel  to  CD,  ABF  is  a  light  angle, 
and  consequently  AF  is  a  diameter,  and  BC  =  BF;  and 
because  AF  is  a  diameter,  J.2>^  is  a  right  angle.  As 
ABB  is  a  right  angle, 

AF'+FD^AD' 

Also, FB'+EC^BC^DF2 

Byadditio^XI^^SV^'+^^ZS'+SF^AF1 
Hence  the  theorem. 


THEOREM  XXIV. 

If  perpendiculars  be  drawn  bisecting  the  three  sides  of  a 
triangle,  they  will,  when  sufficiently  produced,  meet  in  a  com- 
mon point. 

The  three  angular  points  of  a  triangle  are  not  in  the 
same  straight  line;  consequently  one  circumference, 
and  but  one,  may  be  made  to  pass  through  them. 

Conceive  a  triangle  to  be  thus  circumscribed.  The 
sides  of  the  triangle  then  become  chords  of  the  circum- 
scribing circle.  Now  if  these  sides  be  bisected,  and  at  the 
points  of  bisection  perpendiculars  be  drawn  to  the  sides, 
each  of  these  perpendiculars  will  pass  through  the  center 
of  the  circle  (Th.  1,  Cor.) ;  and  the  perpendiculars  will 
therefore  meet  in  a  common  point. 
Hence  the  theorem. 

THEOREM  XXV. 

The  sums  of  the  opposite  sides  of  a  quadrilateral  circum- 
scribing a  circle  are  equal. 

Let  ABOB  be  a  quadrilateral  circumscribed  about  a 
circle,  whose  center  is  0.     Then  we  are  to  prove  that 
AB  +  BC=AB  +  BC. 
From  the  center  of  the  circle  draw  OF  and  OF  to 
the  points  of  contact  of  the  sides  AB  and  BO.     Then, 
10 


ttO 


GEOMETRY. 


the  two  right-angled  triangles,  OEB  and  OFB,  are  equal, 
because  they  have  the  hypotenuse 
OB  common,  and  the  side  OF  = 
OE;  therefore,  BE  =  BF,  (Cor., 
Th.  39,  B.  I). 

In  like  manner  we  can  prove 
that 
AE=  AH,  CF=  Ca,  nn&DG=DK 

Now,  taking  the  equation  BE  = 
BF,  and  adding  to  its  first  mem- 
ber CG-,  and  to  its  second  the 
equal  line  CF.  we  have, 

BE+  CG  =  BF+  OF    (1) 

The  equation  AE=AH,  by  adding  to  its  first  member 
DCr,  and  to  the  second  the  equal  line,  DH,  gives 
AE+DG=AH+DH    (2) 

By  the  addition  of  (1)  and  (2)?  we  find  that 

BE  +  AE+CG  +  DG  =  BF  +  CF+AK+DIT. 

That  is,  AB  +  CD  =  BC  +  AD. 

Hence  the  theorem. 


aooK  IV. 


Ill 


BOOK  IV. 


PROBLEMS 

In  this  section,  we  have,  in  most  instances,  merely 
shown  the  construction  of  the  prohlem,  and  referred  to 
the  theorem  or  theorems  that  the  student  may  use,  to 
prove  that  the  object  is  attained  by  the  construction. 

In  obscure  and  difficult  problems,  however,  we  have 
gone  through  the  demonstration  as  though  it  were  a 
theorem. 

PROBLEM    I. 


To  bisect  a  given  finite  straight  line. 

Let  AB  be  the  given  line,  and  from 
its  extremities,  A  and  B,  with  any 
radius  greater  than  one  half  of  AB, 
(Postulate  3),  describe  arcs,  cutting  A  — 
each  other  in  n  and  m.  Draw  the  line 
nm ;  and  0,  where  it  cuts  AB,  will  be 
the  middle  of  the  given  line. 

Proof,  (B.  I,  Th.  18,  Sch.  2). 

PROBLEM    II. 

To  bisect  a  given  angle. 

Let  ABO  be  the  given  angle.  With  any 
radius,  and  B  as  a  center,  describe  the  arc 
AC.  From  A  and  0,  as  centers,  with  a 
radius  greater  than  one  half  of  AC,  de- 
scribe arcs,  intersecting  in  n;  join  B  and  n; 
the  joining  line  will  bisect  the  given  angle. 

Proof,  (Th.  21,  B.  I). 


X 


X 


112 


GEOMETRY. 


Proof, 


X-tf 


PROBLEM    III. 

From  a  given  point  in  a  given  line,  to  draw  a  perpendicular 
to  that  line. 

Let  AB  be  the  given  line,  and 
0  the  given  point.  Take  n  and  m, 
at  equal  distances  on  opposite  sides 
of  0;  and  with  the  points  m  and 
n,  as  centers,  and  any  radius 
greater  than  nO  or  mO,  describe 
arcs  cutting  each  other  in  &  Draw 
SO,  and  it  will  be  the  perpendicular  required. 
(B.  I,  Th.  18,  Sch.  2). 

The  following  is  another  method, 
which  is  preferable,  when  the  given 
point,  0,  is  at  or  near  the  end  of  the 
line. 

Take  any  point,  0,  which  is  mani- 
festly one  side  of  the  perpendicular, 
as  a  center,  and  with  00  as  a  radius,  describe  a  circum- 
ference, cutting  AB  in  m  and  0.  Draw  mn  through  the 
points  m  and  0,  and  meeting  the  arc  again  in  n ;  mn  is 
then  a  diameter  to  the  circle.  Draw  On,  and  it  will  be 
the  perpendicular  required.     Proof,  (Th.  9,  B.  III). 

PROBLEM   IV. 

From  a  given  point  without  a  line,  to  draw  a  'perpendicular 
to  that  line. 

Let  AB  be  the  given  line,  and  0 
the  given  point.  From  0  draw  any 
oblique  line,  as  On.  Find  the  mid- 
dle point  of  On  by  Problem  1,  and 
with  that  point,  as  a  center,  describe 
a  semicircle,  having  On  as  a  diam- 
eter. From  m,  where  this  semi-cir- 
cumference cuts  AB,  draw  Om,  and  it  will  be  the  pcrpcn 
dicular  required.     Proof,  (Th.  9,  B.  III). 


BOOK  IV. 


113 


PROBLEM   V. 

At  a  given  point  in  a  line,  to  construct  an  angle  equal  to 
a  given  angle. 

Let  A  be  the  point  given  in  the  line 
AB,  and  DOE  the  given  angle. 

With  0  as  a  center,  and  any  radius, 
OE,  draw  the  are  ED. 

With  i  as  a  center,  and  the  radius 
AF=OE,  describe  an  indefinite  arc;  and 
with  J7  as  a  center,  and  FCr  as  a  radius, 
equal  to  ED,  describe  an  arc,  cutting  the 
other  arc  in  Q-,  and  draw  A  G\  QAF  will  be  the  angle 
required.    Proof,  (Th.  2,  B.  III). 


PROBLEM   VI. 

From  a  given  point,  to  draw  a  line  parallel  to  a  given  line. 

Let  A  be  the  given  point,  and  BO  the 
given  line.  Draw  A  0,  making  an  angle, 
AOB;  and  from  the  given  point,  A,  in 
the  line  A  0,  draw  the  angle  OAD  m 
AOB,  by  Problem  5. 

Since  AD  and  BO  make  the  same  angle  with  A 0,  they 
are,  therefore,  parallel,  (B.  I,  Th.  7,  Cor.  1). 


PROBLEM    VII. 
To  divide  a  given  line  into  any  number  of  equal  parte. 

Let  AB  represent  the  given 
line,  and  let  it  be  required  to  di- 
vide it  into  any  number  of  equal 
parts,  say  £ve.  Prom  one  end  of 
the  line  A,  draw  AD,  indefinite 
in  both  length  and  position.  Take 
any  convenient  distance  in  the  di- 
10*  h 


114 


GEOMETRY. 


viders,  as  A  a,  and  set  it  off  on  the  line  AD,  thus  making 
the  parts  Aa,  ab,  be,  etc.,  equal.  Through  the  last  point, 
e,  draw  EB,  and  through  the  points  a,  b,  c,  and  d,  draw 
parallels  to  eB,  by  Problem  6 ;  these  parallels  will  divide 
the  line  a3  required.    Proof,  (Th.  17,  Book  II). 


PROBLEM   VIII. 
To  find  a  third  proportional  to  two  given  lines. 


Let  AB  and  A  0  be  any  two  lines. 
Place  them  at  any  angle,  and  draw 
OB.  On  the  greater  line,  AB,  take 
AD  mm  AC,  and  through  D,  draw 
DE  parallel  to  BO',  AE  is  the  third 
proportional  required. 

Proof,  (Th.  17,  B.  II). 


PROBLEM    IX, 


To  find  a  fourth  proportional  to  three  given  lines. 


L^t  AB,  AC,  AD,  represent  the 
ihrca  given  lines.  Place  the  first 
two  at  any  angle,  as  BAO,  and  draw 
BO  On  AB  place  AD,  and  from 
the  point  D,  draw  DE  parallel  to 
BO,  by  Problem  6 ;  AE  will  be  the 
fourth  proportional  required. 

Proof,  (Th.  17,  B.  H). 


PROBLEM   X. 

Tt  find  the  middle,  m  mean  proportional,  between  two  given 
lines 


BOOK    IV. 


115 


Place  AB  and  BC  in  one  right 
jne,  and  on  A  C,  as  a  diameter,  de- 
scribe a  semicircle,  (Postulate  3), 
and  from  the  point  B,  draw  BD  at 
right  angles  to  AC,  (Problem  3); 
BD  is  the  mean  proportional  re- 
quired. 

Proof,  (B.  m,  Th.  17,  Cor.). 

PROBLEM  XI. 

To  find  the  center  of  a  given  circle. 
Draw  any  two  chords  in  the  given  cir- 
cle, as  AB  and  CD,  and  from  the  middle 
points,  m  and  n,  draw  perpendiculars  to 
AB  and  CD ;  the  point  at  which  these 
two  perpendiculars  intersect  will  be  the 
center  of  the  circle. 

Proof,  (B.  m,  Th.  1,  Cor.). 


PROBLEM  XII. 

To  draw  a  tangent  to  a  given  circle,  from  a  given 
either  in  or  without  the  circumference  of  the  circle* 

When  the  given  point  is  in  the  cir- 
cumference, as  A,  draw  the  radius  A  C, 
and  from  the  point  A,  draw  AB  per- 
pendicular to  AC;  AB  is  the  tangent 
required. 

Proof,  (Th.  4,  B.  III). 

When  the  given  point  is  without 
the  circle,  as  A,  draw  AC  to  the 
center  of  the  circle ;  on  A  C,  as  a 
diameter,  describe  a  semicircle ;  and 
from  B,  where  the  semi-ciruumfer- 
ence  cuts  the  given  circumference, 
draw  AB,  and  it  will  be  tangent  to  the  circle. 

Proof,  (Th.  9,  B.  Ill),  and,  (Th.  4,  B.  III). 


point, 


116 


GEOMETRY. 


PROBLEM    XIII. 

On  a  given  line,  to  describe  a  segment  of  a  circle,  that  shall 
contain  an  angle  equal  to  a  given  angle. 

Let  AB  be  the  given 
line,  and  0  the  given 
angle.  At  the  ends  of 
the  given  line,  form  angles 
DAB,  DBA,  each  equal 
to  the  given  angle,  C. 
Then  draw  AH  and  BE 
perpendiculars  to  AD  and  BD ;  and  with  E  as  a  center, 
and  EA,  or  EB,  as  a  radius,  describe  a  circle;  then  AFB 
will  be  the  segment  required,  as  any  angle  F,  made  in 
it,  will  be  equal  to  the  given  angle,  0. 

Proof,  (Th.  11,  B.  Ill),  and  (Th.  8,  B.  III). 

PROBLEM    XIV. 

From  any  given  circle  to  cut  a  segment,  that  shall  contain 
a  given  angle. 

Let  0  be  the  given  angle.  Take 
any  point,  as  A,  in  the  circumfer- 
ence, and  from  that  point  draw  the 
tangent  AB ;  and  from  the  point 
A,  in  the  line  AB,  construct  the 
angle  BAD  =  0,  (Problem  5),  and  O 
AED  is  the  segment  required. 

Proof,  (Th.  11,  B.  HI),  and  (Th.  8,  B.  HI). 


PROBLEM   XV. 

To  construct  an  equilateral  triangle  on  a  given  straight  line. 

Let  AB  be  the  given  line;  from 
the  extremities  A  and  B,  as  centers, 
with  a  radius  equal  to  AB,  describe  arcs 
cutting  each  other  at  0.  From  0,  the 
point  of  intersection,  draw  QA  and  CB; 
ABO  will  be  the  triangle  required. 

The  construction  is  a  sufficient  demonstration.    Or,  (Ax.  1\ 


BOOK    IV. 


117 


PROBLEM    XVI. 

To  construct  a  triangle,  having  its  three  sides  equal  to  three 
jiven  lines,  any  two  of  which  shall  be  greater  than  the  third. 

Let  AB,  OB,  and  EF,  represent  the       E p 

three  lines.   Take  any  one  of  them,  as  c D 

AB,  to  be  one  side  of  the  triangle.  From 
B,  as  a  center,  with  a  radius  equal  to  OB, 
describe  an  arc ;  and  from  A,  as  a  center, 
with  a  radius  equal  to  EF,  describe  an- 
other arc,  cutting  the  former  in  n.  Draw 
An  and  Bn,  and  AnB  will  be  the  A  re- 
quired.     Proof,  (Ax.  1). 


PROBLEM  XVII. 

To  describe  a  square  on  a  given  line. 
Let  AB  be  the  given  line ;  and  from  the 
extremities,  A  and  B,  draw  A  0  and  BB  per-     9 
pendicular  to  AB.     (Problem  3.) 

From  A,  as  a  center,  with  AB  as  radius, 
strike  an  arc  across  the  perpendicular  at  0;      i 
and  from  0  draw  OB  parallel  to  AB ;  AOBB 
is  the  square  required.      Proof,  (Th.  26,  B.  I). 


PROBLEM  XVIII. 

To  construct  a  rectangle,  or  a  parallelogram,  whose  adia 
.ent  sides  are  equal  to  two  given  lines. 

Let  AB  and  A  0  be  the  two  given        A c 

lines.     From  the  extremities  of  one        A n 


line,  draw  perpendiculars  to  that  line,  as  in  the  last  prob- 
lem; and  from  these  perpendiculars,  cut  off  portions 
equal  to  the  other  line ;  and,  by  a  parallel,  complete  the 
figure. 


118  GEOMETRY. 

When  the  figure  is  to  be  a  parallelogram,  with  oblique 
angles,  describe  the  angles  by  Problem  5.  Proof,  (Th 
26,  B.  I). 

PROBLEM  XIX. 

To  describe  a  rectangle  that  shall  be  equivalent  to  a  given 
square,  and  have  a  side  equal  to  a  given  line. 

Let  AB  be  a  side  of  the  given  square,        c D 

and  CD  one  side  of  the  required  rect-       A B 

angle.  E p 

Find  the  third  proportional,  FF,  to  CD  and  AB,  (Prob- 
lem 8).     Then  we  shall  have 

CD  :  AB  : :  AB  :  FF, 

Construct  a  rectangle  with  the  two  given  lines,  CD 
and  FF,  (Problem  18),  and  it  will  be  equal  to  the  given 
square,  (Th.  3,  B.  II). 

PROBLEM  XX. 

To  construct  a  square  that  shall  be  equivalent  to  the  differ 
ence  of  two  given  squares. 

Let  A  represent  a  side  of  the  greater  of  two  given 
squares,  and  B  a  side  of  the  less  square. 

On  A,  as  a  diameter,  describe  a 
semicircle,  and  from  one  extremity, 
n,  as  a  center,  with  a  radius  equal  to 
B,  describe  an  arc.  and,  from  the 
point  where  it  cuts  the  circumference,  — - — 

draw  mp  and  np  ;  mp  is  the  side  of 
a  square,  which,  when  constructed, 
(Problem  17),  will  be  equal  to  the  difference  of  the  two 
given  squares.     Proof  (Th.  9,  B.  Ill,  and  Th.  39,  B.  I.) 

To  construct  a  square  equivalent  to  the  sum  of  two 
given  squares,  we  have  only  to  draw  through  any  point 
two  lines  at  right  angles,  and  lay  off  on  one  a  distance 
equal  to  the  side  of  one  of  the  squares,  and  on  the  other 


BOOK    IV. 


119 


a  distance  equal  to  the  side  of  the  other.  The  straight 
line  connecting  the  extremities  of  these  lines  will  be  the 
side  of  the  required  square,  (Th.  39,  B.  I). 


PROBLEM  XXI. 

To  divide  a  given  line  into  two  parts,  which  shall  be  in  the 
ratic  of  two  other  given  lines. 


M^ 


NH 


Let  AB  be  the  line  A — ~<B 

to  be  divided,  and  M 
and  N  the  lines  hav- 
ing the  ratio  of  the 
required  parts  of  AB. 
From  the  extremity 
A  draw  AB,  making 
any  angle  with  AB, 
and  take  AC  =  M, 
and  OB  =  N.  Join 
the  points  B  and  B 
by  a  straight  line, 
and  through  (7  draw 
CG  parallel  to  BB. 
Then  will  the  point  G  divide  the  line  AB  into  pa„U 
having  the  required  ratio.  (Proof,  Th.  17,  B.  II). 

Or,  having  drawn  AB,  lay  off  A  0  =  M,  and  through 
B  draw  B  V  parallel  to  AB,  making  it  equal  to  N,  and 
join  0  and  Vbj  a  line  cutting  AB  in  the  point  G. 

Then  the  two  triangles  AOG  and  GrBV  are  equiangu- 
lar and  similar,  and  their  homologous  sides  give  tho 
proportion, 

AG  :  GB  ::  AC  :  BV  ::  M  :  if 

The  line  AB  is  therefore  divided,  at  the  point  G,  into 
parts  which  are  in  the  ratio  of  the  lines  M  and  JV1 


l;*0 


GEOMETRY. 


r> 


PROBLEM  XXII. 

To  divide  a  given  line  into  any  number  of  parts,  having  to 
each  other  the  ratios  of  other  given  lines* 

Let  AB  be  the  given  M»- 
line  to  be  divided,  and  Nl 
M,  It,  P,  etc.,  the  lines 
to  which  the  parts  of 
AB  are  to  be  propor- 
tional. 

Through  the  point  A 
draw  an  indefinite  line,  making,  with  AB,  any  conve- 
nient angle,  and  on  this  line  lay  off  from  A  the  lines  M, 
N,  P,  etc.,  successively.  Join  the  extremity  of  the  last 
line  to  the  point  B  by  a  straight  line,  parallel  to  which 
draw  other  lines  through  the  points  of  division  of  the 
indefinite  line,  and  they  will  divide  the  line  AB  at  the 
points  0,  D,  etc.,  into  the  required  parts.  (Proof,  Th.  17, 
B.  II). 

PROBLEM    XXIII. 

To  construct  a  square  that  shall  be  to  a  given  square,  as  a 
line,  M,  to  a  line,  N. 

Place  M  and  N  in  a  line,  and 
on  the  sum  describe  a  semicir- 
cle. From  the  point  where  the 
two  lines  meet,  draw  a  perpen- 
dicular to  meet  the  circumfer- 
ence in  A.  Draw  Am  and  An, 
and  produce  them  indefinitely.  On  An  or  An  produced, 
take  AG  =  to  the  side  of  the  given  square  ;  and  from  C, 
draw  CB  parallel  to  mn  ;  AB  is  a  side  of  the  required 
square. 


For, 
Also, 


Am 


An* 
An 


Am* 
Therefore,  ~A&  :A02::M 


AB2 :  AC\  (Th.  17,  B.  II). 
M     :JST,      (Th.25,B.II). 
JST,      (Th.  6,  B.  II). 


BOOK    IV.  121 

PROBLEM   XXIV. 

To  cut  a  line  into  extreme  and  mean  ratio  ;  that  is,  sj  that 
the  whole  line  shall  be  to  the  greater  part,  as  that  greater  part 
is  to  the  less. 

Remark.  —  The  geometrical  solution  of  this  problem  is  not  imme- 
diately apparent,  but  it  is  at  once  suggested  by  the  form  of  the  equa- 
tion, which  a  simple  algebraic  analysis  of  its  conditions  leads  to. 

Represent  the  line  to  be  divided  by  2a,  the  greater 
part  by  x,  and  consequently  the  other,  or  less  part,  by 
2a  —  x. 

Now,  the  given  line  and  its  two  parts  are  required,  to 
satisfy  the  following  proportion  : 

2a  :  x  : :  x  :  2a  —  x 

whence,  x%  =  4a2  —  2ax 

By  transposition,     x2  -f  2ax  =  4a2  =  (2a)2 

If  we  add  a2  to  both  members  of  this  equation,  we 
shall  have, 

x2  +  2ax  +  a2  =  (2a)2  -f  a2 

or,  (x  -fa)2  =  (2a)2  -f  a2 

This  last  equation  indicates  that  the  lines  represented 
by  (x  +  a),  2a,  and  a,  are  the  three  sides  of  a  right- 
angled  triangle,  of  which  (x  +  a)  is  the  hypotenuse,  the 
given  line,  2a,  one  of  the  sides,  and  its  half,  a,  the  other. 

Therefore,  let  AB  represent  the 
given  line,  and  from  the  extremity,  B, 
draw  BO  at  right  angles  to  AB,  and 
make  it  equal  to  one  half  of  AB. 

With  C,  as  a  center,  and  radius  OB, 
describe  a  circle.  Draw  A  C  and  pro- 
duce it  to  F.  With  A  as  a  center 
and  AD  as  a  radius,  describe  the  arc 
D.E;  this  arc  will  divide  the  line  AB, 
as  required. 

We  are  now  to  prove  that 

AB  :  AE  ::  AE  :  EB 
11 


122  GEOMETRY. 

By  Th.  18,  B.  Ill,  we  have, 

AF  x  AD  =  AB2 

or,  AF  :  AB  : :  AB  :  AD 

Then,  (by  Cor.,  Th.  8,  Book  II),  we  may  have, 
(AF—  AB)  :  AB  ::  (AB  —  AD)  :  AD 
Since         OB  =  \AB  =  \DF\  therefore,  AB  =  LI 
Hence,  AF — AB  =  AF  —  DF  —  AD  =  A#. 

Therefore,    AF  :  AB  ::  FB  :  AF 
By  taking  the  extremes  for  the  means,  we  have, 
AB  i  AF  ii  AF  :  FB. 

PROBLEM    XXV. 

To  describe  an  isosceles  triangle,  having  its  two  equal  angles 
each  double  the  third  angle,  and  the  equal  sides  of  any  given 
length. 

Let  AB  be  one  of  the  equal  sides  of 
the  required  triangle ;  and  from  the 
point  A,  with  the  radius  AB,  describe 
an  arc,  BD. 

Divide  the  line  AB  into  extreme  acd 
mean  ratio  by  the  last  problem,  and  sup- 
pose 0  the  point  of  division,  and  A  0  the 
greater  segment. 

From  the  point  B,  with  AC,  the  greater  segment,  as  a 
radius,  describe  another  arc,  cutting  the  arc  BD  in  D. 
Draw  BD,  DO,  and  DA.  The  triangle  ABD  is  the  tri- 
angle required. 

As  AC  =  BD,  by  construction ;  and  as  AB  is  to  A 0 
as  A  C  is  to  B C,  by  the  division  of  AB ;  therefore 
AB  :  BD  : :  BD  :  BC 

Now,  as  the  terms  of  this  proportion  are  the  sides  ot" 
the  two  triangles  about  the  common  angle,  B,  it  follows, 
(Cor.  2,  Th.  17,  B.  II),  that  the  two  triangles,  ABD  and 


BOOK   IV. 


123 


BBC,  are  equiangular;  but  the  triangle  ABB  is  isos- 
celes; therefore,  BBQ  is  isosceles  also,  and  BB  ==  BC\ 
but  BB  =  AC:  hence,  BC  =  AC,  (Ax.  1),  and  the  tri- 
angle ACB  is  isosceles,  and  the  [__  CBA  =  | -^L.     But 

the  exterior  angle,  BCB  =  CBA  -f  A,  (Th.  12,  B.  I), 
Therefore,  \__BCB,  or  its  equal  [__B  =  L CBA  +  [__A ;  or 
the  angle  B  =  2[__A.  Hence,  the  triangle  ABB  has  each 
of  its  angles,  at  the  base,  double  of  the  third  angle. 

Scholium. — As  the  two  angles,  at  the  base  of  the  triangle  ABD,  are 
equal,  and  each  is  double  the  angle  A,  it  follows  that  the  sum  of  the 
three  angles  is  Jive  times  the  angle  A.  But,  as  the  three  angles  of  every 
triangle  are  always  equal  to  two  right  angles,  or  180°,  the  angle  A 
must  be  one  fifth  of  two  right  angles,  or  36° ;  therefore,  BD  is  a  chord 
of  36°,  when  AB  is  a  radius  to  the  circle ;  and  ten  such  chords  would 
extend  exactly  round  the  circle,  or  would  form  a  decagon. 


PROBLEM  XXVI, 


Within  a  given  circle  to  inscribe  a  triangle,  equiangular  to 
a  given  triangle. 

Let  ABC  be  the  circle,  and 
abc  the  given  triangle.  From 
any  point,  as  A,  draw  EB  tan- 
gent to  the  given  circle  at  A9 
(Problem  12). 

From  the  point  A,  in  the  line 
AB,  lay  off  the  angle  BAC  — 
the  angle  b,  (Problem  5),  and  the  angle  BAB  —  the  angle 
c,  and  draw  BC. 

The  triangle  ABC  is  inscribed  in  the  circle;  It  13  equi- 
angular to  the  triangle  abc,  and  hence  it  is  the  triangle 
required. 

Proof,  (Th.  12,  B.  HI). 


124 


GEOMETRY. 


PROBLEM   XXVII, 


To  inscribe  a  regular  pentagon  in  a  given  circle. 

1st.  Describe  an  isosceles  tri- 
angle, abc,  having  each  of  the 
equal  angles,  b  and  c,  double  the 
third  angle,  a,  by  Problem  25. 

2d.    Inscribe    the     triangle, 
ABO,  in  the  given  circle,  equi- 
angular to  the  triangle  abc,  by 
Problem  26  ;  then  each  of  the  angles,  B  and  0,  is  double 
the  angle  A. 

3d.  Bisect  the  angles  B  and  0,  by  the  lines  BD  and 
OE,  (Problem  2),  and  draw  AE,  EB,  OB,  BA;  and  the 
figure  AEBOB  is  the  pentagon  required. 

By  construction,  the  angles  BAO,  ABB,  BBO,  BOE, 
EOA,  are  all  equal ;  therefore,  (B.  in,  Th.  9,  Cor.),  the 
arcs,  BO,  AB,  BO,  AE,  and  EB,  are  all  equal;  and  if 
the  arcs  are  equal,  the  chords  AE,  EB,  etc.,  are  equal. 

Scholium. — The  arc  subtended  by  one  of  the  sides  of  a  regular  pen- 

360° 
tagon,  being  one  fifth  of  the  whole  circumference,  is  equal  to  — — ==72°' 

u 


PROBLEM   XXVIII. 

To  inscribe  a  regular  hexagon  in  a  circle. 

D  aw  any  diameter  of  the  circle,  as 
AB.  and  from  one  extremity,  B,  draw 
BB  equal  to  BO,  the  radius  of  the 
circle.  The  arc,  BB,  will  be  one  sixth 
part  of  the  whole  circumference,  and 
the  chord  BB  will  be  a  side  of  the  regu- 
lar polygon  of  six  sides. 

In  the  A  OBB,  as   OB  =  OB,  and  BB=  OB  by  con- 
struction,  the  A  is  equilateral,  and  of  course  equiangular. 

Since  the  sum  of  the  three  angles  of  every  A  is  equal 
to  two   right    angles,   or    to    180    degrees,   when   the 


BOOK   IV.  125 

three  angles  are  equal  to  one  another,  each  one  of  them 
must  be  60  degrees ;  but  60  degrees  is  a  sixth  part  of 
360  degrees,  the  whole  number  of  degrees  in  a  circle ; 
therefore,  the  arc  whose  chord  is  equal  to  the  radius,  is  a 
sixth  part  of  the  circumference ;  and,  if  a  polygon  of  six 
equal  sides  be  inscribed  in  a  circle,  each  side  will  be 
equal  to  the  radius. 

Scholium.  —  Hence,  as  BD  is  the  chord  of  60°,  and  equal  to  BC  or 
CD,  we  say  generally,  that  the  chord  of  60°  is  equal  to  radius. 

PROBLEM  XXIX. 

To  find  the  side  of  a  regular  polygon  of  fifteen  sides,  which 
may  be  inscribed  in  any  given  circle. 

Let  CB  be  the  radius  of  the  given 
circle;  divide  it  into  extreme  and 
mean  ratio,  (Problem  24),  and  make 
BD  equal  to  CB,  the  greater  part; 
then  BD  will  be  a  side  of  a  regular 
polygon  of  ten  sides,  (Scholium  to 
Problem  25).  Draw  BA  =  to  CB,  and 
it  will  be  a  side  of  a  polygon  of  six  sides.  Draw  DA, 
and  that  line  must  be  the  side  of  a  polygon  which  cor- 
responds to  the  arc  of  the  circle  expressed  by  \  less  j\, 
of  the  whole  circumference ;  or  J  —  j^  =  g%  =  T^ ;  that 
is,  one-fifteenth  of  the  whole  circumference ;  or,  DA  is 
a  side  of  a  regular  polygon  of  15  sides.  But  the  15th 
part  of  360°  is  24°  ;  hence  the  side  of  a  regular  inscrioed 
polygon  of  fifteen  sides  is  the  chord  of  an  arc  of  24°. 


PROBLEM    XXX. 

In  a  given  circle  to  inscribe  a  regular  polygon  of  any  num 
her  of  sides,  and  then  to  circumscribe  the  circle  by  a  similar 
polygon. 

11* 


126 


GEOMETKY. 


Let  the  circumference  of  the  circle,  whose  cetter  is  C, 
be  divided  into  any  number  of  equal  arcs,  as    amb,   bnc, 
cod,  etc.  ;    then  will  the  polygon  abode,  etc.,  bounded  by 
the  chords  of  these  arcs,  be  regu- 
lar and  inscribed ;  and  the  poly- 
gon ABODE,  etc.,  bounded  by 
the  tangents  to  these  arcs  at  their 
middle  points  m,  n,  o,  etc,  be  a 
3imilar  circumscribed  polygon. 

First.  —  The  polygon  abode, 
etc.,  is  equilateral,  because  its 
sides   are  the  chords  of  equal 

arcs  of  the  same  circle,  (Th.  5,  B.  Ill) ;  and  it  is  equi- 
angular, because  its  angles  are  inscribed  in  equal  segments 
of  the  same  circle,  (Th.  8,  B.  III).  Therefore  the  poly- 
gon is  regular,  (Def.  14,  B.  Ill),  and  it  is  inscribed,  since 
the  vertices  of  all  its  angles  are  in  the  circumference  of 
the  circle,  (Def.  13,  B.  III). 

Second. — If  we  draw  the  radius  to  the  point  of  tangency 
of  the  side  AB  of  the  circumscribed  polygon,  this  radius 
is  perpendicular  to  AB,  (Th.  4,  B.  Ill),  and  also  to  the 
chord  ab,  (B.  Ill,  Th.  1,  Cor.) ;  hence  AB  is  parallel  to  a5, 
and  for  the  same  reason  BO  is  parallel  to  be ;  therefore 
the  angle  ABO  is  equal  to  the  angle  abc,  (Th.  8,  B.  I). 
In  like  manner  we  may  prove  the  other  angles  of  the 
circumscribed  polygon,  each  equal  to  the  corresponding 
angle  of  the  inscribed  polygon.  These  polygons  are 
therefore  mutually  equiangular. 

Again,  if  we  draw  the  radii  Om  and  On,  and  the  line  0B> 
the  two  A's  thus  formed  are  right-angled,  the  one  at  n 
and  the  other  at  n,  the  side  OB  is  common  and  Om  is 
equal  to  On ;  hence  the  difference  of  the  squares  descril^ed 
on  OB  and  Om  is  equivalent  to  the  difference  of  the 
squares  described  on  OB  and  On.  But  the  first  difference 
is  equivalent  to  the  square  described  on  Bm,  and  the 
second  diffeience  is  equivalent  to  the  square  described 


BOOK    IV.  127 

on  Bn ;  hence  Brn  is  equal  to  Bn,  and  the  two  right- 
angled  triangles  are  equal,  (Th.  21,  B.  I),  the  angle  BOm 
opposite  the  side  Bm  being  equal  to  the  angle  BOn,  op- 
posite the  equal  side  Bn.  The  line  OB  therefore  passea 
through  the  middle  point  of  the  arc  mbn ;  hut  because  m 
and  n  are  the  middle  points  of  the  equal  arcs  amb  and 
bnc,  the  vertex  of  the  angle  abc  is  also  at  the  middle 
point  of  the  arc  mbn.  Hence  the  line  OB,  drawn  from 
the  center  of  the  circle  to  the  vertex  of  the  angle  ABC, 
also  passes  through  the  vertex  of  the  angle  abc.  By  pre- 
cisely the  same  process  of  reasoning,  we  may  prove  that 
00  passes  through  the  point  e,  OB  through  the  point  d, 
etc. ;  hence  the  lines  joining  the  center  with  the  vertices 
of  the  angles  of  the  circumscribed  polygon,  pass  through 
the  vertices  of  the  corresponding  angles  of  the  inscribed 
polygon ;  and  conversely,  the  radii  drawn  to  the  vertices 
of  the  angles  of  the  inscribed  polygon,  when  produced, 
pass  through  the  vertices  of  the  corresponding  angles 
of  the  circumscribed  polygon. 

Now,  since  ab  is  parallel  to  AB,  the  similar  A's  abO 
and  ABO,  give  the  proportion 

Ob  :  OB  ::  ab  :  AB, 

and  the  A'a,bcO  and  BOO,  give  the  proportion 

Ob  :  OB  : :  be  :  BO. 

As  these  two  proportions  have  an  antecedent  and  con- 
sequeut,  the  same  in  both,  we  have,  (Th.  6,  B.  II), 

ab  :  AB  : :  bo  :  BO. 

In  like  manner  we  may  prove  that 

be  :  BO  : :  cd  :  OB,  etc.,  etc. 

The  two  polygons  are  therefore  not  only  equiangular, 
but  the  sides  about  the  equal  angles,  taken  in  the  same 
order,  are  proportional ;  they  are  therefore  similar,  (Def 
16.  B.  IT). 


J28  GEOMETRY. 

Cor.  1.  To  inscribe  any  regular  polygon  in  a  circle,  we 
have  only  to  divide  the  circumference  into  as  many  equal 
parts  as  the  polygon  is  to  have  sides,  and  to  draw  the 
chords  of  the  arcs;  hence,  in  a  given  circle,  it  is  possible 
to  inscribe  regular  polygons  of  any  number  of  sides 
whatever.  Having  constructed  any  such  polygon  in  a 
given  circle,  it  is  evident,  that  by  changing  the  radius  of 
the  circle  without  changing  the  number  of  sides  of  the 
polygon,  it  may  be  made  to  represent  any  regular  poly- 
gon of  the  same  name,  and  it  will  still  be  inscribed  in  a 
circle.  As  this  reasoning  is  applicable  to  regular  poly- 
gons of  whatever  number  of  sides,  it  follows,  that  any 
regular  polygon  may  be  circumscribed  by  the  circumference 
of  a  circle. 

Cor.  2.  Since  ab,  be,  cd,  etc.,  are  equal  chords  of  the 
same  circle,  they  are  at  the  same  distance  from  the 
center,  (Th.  3,  B.  Ill) ;  hence,  if  with  0  as  a  center,  and 
Ot,  the  distance  of  one  of  these  chords  from  that  point, 
as  a  radius,  a  circumference  be  described,  it  will  touch 
all  of  these  chords  at  their  middle  points.  It  follows, 
therefore,  that  a  circle  may  be  inscribed  within  any  regular 
polygon. 

Scholium. — The  center,  0,  of  the  circle,  may  be  taken  as  the  center 
of  both  the  inscribed  and  circumscribed  polygons ;  and  the  angle 
A  OB,  included  between  lines  drawn  from  the  center  to  the  extremities 
of  one  of  the  sides  A B,  is  called  the  angle  at  the  center."  The  perpen- 
dicular drawn  from  the  center  to  one  of  the  sides  is  called  the  Ajpothem 
of  the  polygon. 

Cor.  3.  The  angle  at  the  center  of  any  regular  polygon 
is  equal  to  four  right  angles  divided  by  the  number  of 
sides  of  the  polygon.  Thus,  if  n  be  the  number  of  sides 
of  the  polygon,  the  angle  at  the  center  will  be  expressed 
v  360° 
n 

Cor.  4.  If  the  arcs  subtended  by  the  sides  of  any 
regular  inscribed  polygon  be  bisected,  and  the  chords 
of  these  semi-arcs  be  drawn,  we  shall  have  a  regular 


BOOK   IV.  129 

inscribed  polygon  of  double  the  number  of  sides.  Thus, 
from  the  square  we  may  pass  successively  to  regular 
inscribed  polygons  of  8,  16,  32,  etc.,  sides.  To  get  the 
corresponding  circumscribed  polygons,  we  have  merely 
to  draw  tangents  at  the  middle  points  of  the  arcs  sub- 
tended by  the  sides  of  the  inscribed  polygons. 

Cor,  5.  It  is  plain  that  each  inscribed  polygon  is  but 
a  part  of  one  having  twice  the  number  of  sides,  while 
each  circumscribed  polygon  is  but  a  part  of  one  having 
o^p  hqlf  thp.  number  of  sides 


130 


GEOMETRY. 


BOOK  V. 


ON  THE   PROPORTIONALITIES  AND  MEASUREMENT 
OF  POLYGONS  AND  CIRCLES. 


PROPOSITION  I.  — THEOREM. 

The  area  of  any  circle  is  equal  to  the  product  of  its  radius 
by  one  half  of  its  circumference. 

Let  CA  be  the  radius  of  a  circle, 
and  AB  a  very  small  portion  of  its 
circumference;  then  AOB  will  be  a 
sector.  We  may  conceive  the  whole 
circle  made  up  of  a  great  number  of 
such  sectors;  and  when  each  sector 
is  very  small,  the  arcs  AB,  BB,  etc., 
each  one  taken  separately,  may  be  regarded  as  nght 
lines  ;  and  the  sectors  CAB,  CBD,  etc.,  will  be  triangles. 
The  triangle,  AOB,  is  measured  by  the  product  of  the 
base,  A 0,  multiplied  into  one  half  the  altitude,  AB,  (Th. 
33,  Book  I) ;  and  the  triangle  BCD  is  measured  by  the  pro- 
duct of  BC,  or  its  equal,  AC,  into  one  half  BB ;  then  the 
area,  or  measure  of  the  two  triangles,  or  sectors,  is  the 
product  of  A  C,  multiplied  by  one  half  of  AB  plus  one 
half  of  BB,  and  so  on  for  all  the  sectors  that  compose 
the  circle ;  therefore,  the  area  of  the  circle  is  measured 
by  th  g  product  of  the  radius  into  one  half  the  circumference. 


BOOK    V.  131 

PROPOSITION  II.  —  THEOREM. 

Circumferences  of  circles  are  to  one  another  as  their  radii, 
and  their  areas  are  to  one  another  as  the  squares  of  their 
radii. 

Let  CA  be  the  radius  of  a  circle, 
and  Ca  the  radius  of  another  circle. 
Conceive  the  two  circles  <x>  be  so 
placed  upon  each  other  so  as  to  have 
a  common  center. 

Let  AB  be  such  a  certain  definite 
portion  of  the  circumference  of  the 
larger  circle,  that  m  times  AB  will  represent  that  cir- 
cumference. 

But  whatever  part  AB  is  of  the  greater  circumference, 
the  same  part  ab  is  of  the  smaller;  for  the  two  circles 
have  the  same  number  of  degrees,  and  are  of  course  sus- 
ceptible of  division  into  the  same  number  of  sectors. 
But  by  proportional  triangles  we  have, 
CA  :  Ca  : :  AB  :  ab 

Multiply  the  last  couplet  by  m,  (Th.  4,  B.  II),  and  we 
have 

CA  :  Ca  ::  m.AB  :  m.ab. 

That  is,  the  radius  of  one  circle  is  to  the  radius  of  another, 
as  the  circumference  of  the  one  is  to  the  circumference  of  the 
other. 

To  prove  the  second  part  of  the  theorem,  let  C  repre- 
sent the  area  of  the  larger  circle,  and  c  that  of  the 
smaller ;  now,  whatever  part  the  sector  CAB  is  of  the 
circle  C,  the  sector  Cab  is  the  corresponding  part  of  the 
circle  c. 


That  is,  C :  c 

but,  CAB  :  Cab 

Therefore,         C :  c 


CAB  :  Cab, 

{CAf:(Ca)2,    (Th.20,B.H). 

(CAY  :  (Caf,    (Th.  6,  B.  II). 


That  is,  the  area  of  one  circle  is  to  the  area  of  another,  as 


132  GEOMETRY. 

the  square  of  the  radius  of  the  one  is  to  the  square  of  the 
radius  of  the  other. 
Hence  the  theorem. 

Cor.     If  0  :  c  ::  (OAf    :  (Ca)2, 

then,  C  :  c  ::4(£M)2  :  4  ((7a)2. 

But  4  (QAf  is  the  square  of  the  diameter  of  the  larger 
circle,  and  4  (Oaf  is  the  square  of  the  diameter  of  the 
smaller.  Denoting  these  diameters  respectively  by  B 
and  d,  we  have, 

0  :  c  : :  D2  :  d\ 

That  is,  the  areas  of  any  two  circles  are  to  each  other,  a% 
the  squares  of  their  diameters. 

Scholium.  —  As  the  circumference  of  every  circle,  great  or  small,  is 
assumed  to  be  the  measure  of  360  degrees,  if  we  conceive  the  circum- 
ference to  be  divided  into  360  equal  parts,  and  one  such  part  repre- 
sented by  AB  on  one  circle,  or  ab  on  the  other,  AB  and  ab  will  be  very 
near  straight  lines,  and  the  length  of  such  a  line  as  ^LBwill  be  greater 
or  less,  according  to  the  radius  of  the  circle ;  but  its  absolute  length 
cannot  be  determined  until  we  know  the  absolute  relation  between  th« 
diameter  of  a  circle  and  its  circumference. 

PROPOSITION    III.  — THEOREM. 

When  the  radius  of  a  circle  is  unity,  its  area  and  semi- 
circumference  are  numerically  equal. 

Let  B  represent  the  radius  of  any  circle,  and  the  Greek 
letter,  *,  the  half  circumference  of  a  circle  whose  radius 
is  unity.  Since  circumferences  are  to  each  other  as  their 
radii,  when  the  radius  is  B,  the  semi-circumference  will 
be  expressed  by  *B. 

Let  m  denote  the  area  of  the  circle  of  which  B  is  the 
radius ;  then,  by  Theorem  1,  we  shall  have,  for  the  area 
of  this  circle,  icB2  =  my  which,  when  B  =  1,  reduces  to 
«r  =  m. 

This  equation  is  to  be  interpreted  as  meaning  that  the 
semi-circumference  contains  its  unit,  the  radius,  as  many 


BOOK  V.  133 

times  as  the  area  of  the  circle  contains  its  unit,  the 
square  of  the  radius. 

Remark.  —  The  celebrated  problem  of  squaring  the  circle  has  for  its 
object  to  find  a  line,  the  square  on  which  will  be  equivalent  to  the  area 
of  a  circle  of  a  given  diameter ;  or,  in  other  words,  it  proposes  to  find 
the  ratio  between  the  area  of  a  circle  and  the  square  of  its  radius. 

An  approximate  solution  only  of  this  problem  has  been  as  yet  dis- 
covered, but  the  approximation  is  so  close  that  the  exact  solution  ii 
no  longer  a  question  of  any  practical  importance. 


PROPOSITION   IV.  — PROBLEM. 

Given,  the  radius  of  a  circle  unity,  to  find  the  areas  of 
regular  inscribed  and  circumscribed  hexagons. 

Conceive  a  circle  described  with  the  radius  CA,  and  in 
this  circle  inscribe  a  regular  polygon  of  six  sides  (Prob. 
28,  B.  IV),  and  each  side  will  be 
equal  to  the  radius  CA ;  hence, 
the  whole  perimeter  of  this  poly- 
gon must  be  six  times  the  ra- 
dius of  the  circle,  or  three  times 
the. diameter.     The  chord  bd  is  Ft 

bisected  by  CA.  Produce  Cb  and  Cd,  and  through  the 
point  A,  draw  BD  parallel  to  bd ;  BD  will  then  be  a  3ide 
of  a  regular  polygon  of  six  sides,  circumscribed  about 
the  circle,  and  we  can  compute  the  length  of  this  line, 
BD,  as  follows :  The  two  triangles,  Cbd  and  CBD,  are 
equiangular,  by  construction ;  therefore, 

Ca  :  bd  : :  CA  :  BD. 

Now,  let  us  assume  CA  =  Cd  =  the  radius  of  the 
circle,  equal  unity;  then  bd  =  l,  and  the  preceding  pro- 
portion becomes 

Ca  :  1  ::  1  :  BD        (1) 

In  the  right-angled  triangle  Cad,  we  have, 

(Cay  +  (ad)*  =  (C'dy,        (Th.  39,  B.  I). 
That  is,       (Ca)2  +  i  =  l,  because  Cd=l,  and  ad—  \. 
12 


134  GEOMETRY. 

"Whence,  Ca  =  J  >/3.  This  value  of  Ca,  substituted  in 
proportion  ( 1 ),  gives 

|^3  :  1  : :  1  :  BD;  hence,  BD=  JL 

But  the  area  of  the  triangle  Cbd  is  equal  to  bd(=  1,) 
multiplied  by  J  Ca  =  \  ^3 ;  and  the  area  of  the  triangle 
CBD  is  equal  to  BD  multiplied  by  \CA. 

"Whence,         area,  Cbd  =  J  v/3, 

and,  area,  CBD  ■»    ,i 

But  the  area  of  the  inscribed  polygon  is  six  times  that 
of  the  triangle  Cbd,  and  the  area  of  the  circumscribed 
polygon  is  six  times  that  of  the  triangle  CBD. 

Let  the  area  of  the  inscribed  polygon  be  represented 
by  p,  and  that  of  the  circumscribed  polygon  by  P. 

Then^  =  ?V3,  andP ?«  ^J  -  2^3. 

2  •a        v/3 

3  3 

Whence^  :  P  : :  ^3  :  2%/3  ::£:2::3:4::9  :  12 

^  =  2^3  =  2.59807621.    P  =  2^3  =  3.46410161. 

Now,  it  is  obvious  that  the  area  of  the  circle  must  be 
included  between  the  areas  of  these  two  polygons,  and 
not  far  from,  but  somewhat  greater  than,  their  half  sum, 
which  is  3.03  -f  ;  and  this  may 'be  regarded  as  the  first 
approximate  value  of  the  area  of  the  circle  to  the  radius 
unity. 

PROPOSITION   Y.  — PROBLEM. 

(riven,  the  areas  of  two  regular  polygons  of  the  same  nvm 
her  of  sides,  the  one  inscribed  in  and  the  other  circumscribed 
about,  the  same  circle,  to  find  the  areas  of  regular  inscribed  and 
circumscribed  polygons  of  double  the  number  of  sides. 

Let  j?  represent  the  area  of  the  given  inscribed  polygon, 
and  P  that  of  the  circumscribed  polygon  of  the  same 


BOOK   V. 


135 


nunibei  of  sides.  Also  denote  by  p'  the  area  of  the 
inscribed  polygon  of  double  the  number  of  sides,  and  by 
P'  that  of  the  corresponding  circumscribed  polygon. 
Now,  if  the  arc  KAL  be  some  exact  part,  as  one-fourth, 
one  fifth,  etc.,  of  the  circumference  of  the  circle,  of  which 
0  is  the  center  and  CA  the  radius,  then  will  KL  be  the 
side  of  a  regular  inscribed  polygon,  and  the  triangle 
KCL  will  be  the  same  part  of  the  whole  polygon  that 
the  arc  KAL  is  of  the  whole  circumference,  and  the 
triangle  CDB  will  be  a  like  part  of  the  circumscribed 
polygon.  Draw  QA  to  the  point  of  tangency,  and  bisect 
the  angles  ACB  and  ACD,  by  the  lines  CGr  and  CLJ,  and 
draw  KA. 

It  is  plain  that  the  triangle 
ACK  is  an  exact  part  of  the 
inscribed  polygon  of  double  the 
number  of  sides,  and  that  the 
A  ECG-  is  a  like  part  of  the  cir- 
cumscribed polygon  of  double 
the  number  of  sides.  Repre- 
sent the  area  of  the  A  LCKbj 
a,  and  the  area  of  the  A  BCD 
by  b,  that  of  the  A  ACK  by  x, 

and  that  of  the  A  ECG-  by  y,  and  suppose  the  A's,  KCL 
and  BBC,  to  be  each  the  nth.  part  of  their  respective 
polygons. 

Then,  na—p,     nb  =  P,     2nx  =  p', 

and,  2ny  =  Pf ; 

But,  by  (Th.  33,  B.  I),  we  have 

CM.  MK=a  (1) 
CA  .  AD  =b  (2) 
QA  .  MK=  2x      (3) 

Multiplying  equations  (1)  and  (2),  member  by  member, 
we  have 

(CM  .  AD)  x  (CA  .  MK)  =-  ab        (4 ) 


136  GEOMETRY. 

From  the  similar  A's  OMK and  CAD,  we  have 

CM  :  MK  ::  CA  :  AD 
whence  CM  .  AD  =  CA  .  MK 

But  from  equation  ( 3 )  we  see  that  each  memher  of 
this  last  equation  is  equal  to  2x;  hence  equation  (4) 
becomes 

2x  .  2x  =  <*5 

If  we  multiply  both  members  of  this  by  n2  ■■  n  h, 
we  shall  have 

4t&2.z2  =  wa.w5  =  p.P 

or,  taking  the  square  root  of  both  members, 

2nx  =  s/p^P 

That  is,  the  area  of  the  inscribed  polygon  of  double  the 
number  of  sides  is  a  mean  proportional  between  the  areas  of 
the  given  inscribed  and  circumscribed  polygons  p  and  P. 

Again,  since  CE  bisects  the  angle  ACD,  we  have,  by, 
(Th.  24,  B.  II), 


AE  :  ED 


CA  :  CD 
CM:  CK 
CM:  CA 
CM:  CM+  CA. 


hence,    AE  :  AE  +ED 

Multiplying  the  first  couplet  of  this  proportion  by  CA, 
and  the  second  by  MK,  observing  that  AE  -f  ED  =  AD, 
we  shall  have 

AE.CA  :  AD.CA  ::  CM.MK  :  (CM  +  CA)  MK. 

But  AE.  CA  measures  the  area  of  the  A  CEG-,  whi  3h 
we  have  called  y,  AD.CA  =  A  CBD  —  b,  CM.MK  - 
A  CKL  =  a,  and  (CM  4-  CA)MK=  a  CKL  +  2  a  CAK  = 
a  -h  2x,  as  is  seen  from  equations  (1)  and  (3).  Therefore, 
the  above  proportion  becomes 

y  :  b  ::  a  :  a  +  2x. 

Multiplying  the  first  couplet  by  2n,  and  the  second  by 
w,  we  shall  have 


BOOK   V.  137 

2m/  :  2nb  : :  na  :  na  -f  2wa: 
That  is,  P'  :  2P   ::  p    :  p  +  / 

whence,  P'  =  — -±- 

and  as  the  value  of  pr  has  been  previously  fourd  equal  to 
*SPp,  the  value  of  P'  is  known  from  this  last  equation, 
and  the  problem  is  completely  solved. 

PROPOSITION    VI.  — PROBLEM. 

To  determine  the  approximate  numerical  value  of  the  area 
of  a  circle,  when  the  radius  is  unity. 

"We  have  now  found,  (Prob.  4),  the  areas  of  regular 
inscribed  and  circumscribed  hexagons,  when  the  radius 
of  the  circle  is  taken  as  the  unit ;  and  Prob.  5  gives  us 
formulae  for  computing  from  these  the  areas  of  regular 
inscribed  and  circumscribed  polygons  of  twelve  sides, 
and  from  these  last  we  may  pass  to  polygons  of 
twenty-four  sides,  and  so  on,  without  limit.  Now,  it  is 
evident  that,  as  the  number  of  sides  of  the  inscribed 
polygon  is  increased,  the  polygon  itself  will  increase, 
gradually  approaching  the  circle,  which  it  can  never  sur- 
pass. And  it  is  equally  evident  that,  as  the  number  of 
sides  of  the  circumscribed  polygon  is  increased,  the  poly- 
gon itself  will  decrease,  gradually  approaching  the  circle, 
less  than  which  it  can  never  become. 

The  circle  being  included  between  any  two  corres- 
ponding inscribed  and  circumscribed  polygons,  it  will 
differ  from  either  less  than  they  differ  from  each  other ; 
and  the  area  of  either  polygon  may  then  be  taken  as  tne 
area  of  the  circle,  from  which  it  will  differ  by  an  amount 
less  than  the  difference  between  the  polygons. 

It  is  also  plain  that,  as  the  areas  of  the  polygons  ap- 
proach equality,  their  perimeters  will  approach  coinci- 
dence with  each  other,  and  with  the  circumference  of 
the  circle. 
12* 


138  GEOMETRY. 

Assuming  the  areas  already  found  for  the  inscribed 
and  circumscribed  hexagons,  and  applying  the  formulae 
of  Prob.  5  to  them  and  to  the  successive  results  ob. 
tained,  we  may  construct  the  following  table : 


NUMBER  OB 

1   SIDES. 

n 

INSCRIBED  POLYGONS. 

CIRCUMSCRIBED   POLYGONS. 

6 

1^ 

-  2.59807621 

2^3= 

=3.46410161 

12 

24 

3 
6 

=  3.0000000 
=  3.1058286 

12 
2+/3 

-3.2153904 
3.1596602 

^2+x/3 

48 

3.1326287 

3.1460863 

96 

3.1393554 

3.1427106 

192 

3.1410328 

3.1418712 

384 

3.1414519 

3.1416616 

768 

3.1415568 

3.1416092 

1536 

3.1415829 

3.1415963 

3072 

3.1415895 

3.1415929 

6144 

3.1415912 

3.1415927 

Thus  we  have  found,  that  when  the  radius  of  a  circle  is 
1,  the  semi-circumference  must  be  more  than  3.1415912, 
and  less  than  3.1415927;  and  this  is  as  accurate  as  can 
be  determined  with  the  small  number  of  decimals  here 
used.  To  be  more  accurate  we  must  have  more  decimal 
places,  and  go  through  a  very  tedious  mechanical  opera- 
tion; but  this  is  not  necessary,  for  the  result  is  well 
known,  and  is  3.1415926535897,  plus  other  decimal  places 
to  the  100th,  without  termination.  This  result  was  dis- 
covered through  the  aid  of  an  infinite  series  in  the  Dif- 
ferential and  Integral  Calculus. 

The  number,  3.1416,  is  the  one  generally  usod  in  prac- 
tice, as  it  is  much  more  convenient  than  a  greater  num- 
ber of  decimals,  and  it  is  sufficiently  accurate  for  all 
ordinary  purposes. 

In  analytical  expressions  it  has  become  a  general  cus- 
tom with  mathematicians  to  represent  this  number  by 


BOOK  V.  139 

the  Greek  letter  #,  and,  therefore,  when  any  diameter  of 
a  circle  is  represented  by  D,  the  circumference  of  the 
same  circle  must  be  *D.  If  the  radius  of  a  circle  is  re- 
presented by  B,  the  circumference  must  be  represented 
by  2«B. 

Scholium.  —  The  side  of  a  regular  inscribed  hexagon  subtends  an 
arc  of  60°,  and  the  side  of  a  regular  polygon  of  twelve  sides  subtends 
an  arc  of  30° ;  and  so  on,  the  length  of  the  arc  subtended  by  the  sides 
of  the  polygons,  varying  inversely  with  the  number  of  sides. 

Angles  are  measured  by  the  arcs  of  circles  included  between  their 
sides  ;  they  may  also  be  measured  by  the  chords  of  these  arcs,  or  rather 
by  the  half  chords  called  sines  in  Trigonometry.  For  this  purpose,  it 
becomes  necessary  to  know  the  length  of  the  chord  of  every  possible 
arc  of  a  circle. 

PROPOSITION  VII.  —  PROBLEM. 

Given,  the  chord  of  any  arc,  to  find  the  chord  of  one  half 
that  arc,  the  radius  of  the  circle  being  unity. 

Let  FE  be  the  given  chord,  and  draw 
the  radii  QA  and  QE,  the  first  perpen- 
dicular to  FE,  and  the  second  to  its  ex- 
tremity, E. 

Denote  FE  by  2c,  and  the  chord  of 
the  half  arc  AE  by. a;. 

Then,   in   the  right-angled   triangle, 
DOE,   we  have  ~DQ2  =  QE2  —  DE2.    Whence,    since 
QE  =  1,  D<7  =  ^l  —  c\ 


If  from  QA  =■  1  we  subtract  DO,  we  shall  have  AD. 
That  is,  AD  =  1  —  </T=7;  but  AD2  +  DE2  =  AE\ 
and  AD*  =  2  —  2^1  —  c2  —  c\  Adding  to  the  first 
member  of  this  last  equation  DE2,  and  to  the  second  its 
value  c2,  we  have 

AD2  +  DE2=2-2  Vl^7. 

Whence,        AE  =x/2  —  2<Sl  —  c2,  the  value  sought. 

By  applying  this  formula  successively  to  any  known 
chord,  we  can  find  the  chord  of  one  half  the  arc,  that  of 
half  of  the  half,  and  so  on,  to  the  chords  of  the  most 
minute  arcs. 


140  GEOMETRY. 

Application, 

The  greatest  chord  in  a  circle  is  its  diameter,  which  is 
2  when  the  radius  is  1;  therefore,  we  may  commence 
by  making  2c  =  2,  and  c-  =  1. 

Then,  AE  =  ^2-21/T^c8  =  v/2—2v'  l^T  -  ^2  - 
1.41421356,  which  is  the  chord  of  90°. 

ISTow  make  2c  =  1.41421356,  and  c  -.70710678  -  J^X 

We  shall  then  have, 

chord  of  45°= V2- V2=  i/2- 1.41421356-  ^.58578644- 
.7653  +  . 

Again,  placing  2<?=.7653-f-,  and  applying  the  formula, 
we  can  obtain  the  chord  of  22°  30',  and  from  this  the 
chord  of  11°  15',  and  so  on,  as  far  as  we  please. 

We  may  take,  for  another  starting  point,  the  chord  of 
60°,  which  is  known  to  be  equal  to  the  radius  of  the 
circle,(Prob.  26,  B.  IV).  If,  as  above,  we  make  successive 
applications  of  the  formula,  putting  first  2c  =  1,  we  shall 
arrive  at  the  results  in  the  following 


TABLE 

• 

Chord  of  60°, 

=  J  of  a 

circumference,  1.0000000000 

k 

«   30°, 

_  l  a 

M 

.5176380902 

a 

"  15°, 

l  (( 

—    2* 

a 

.2610523842 

u 

"  7°  30', 

1    u 

—    35 

a 

.1308062583 

U 

"  3°  45', 

1   « 

—  -S3 

a 

.0654381655 

u 

"  1°  52'  30", 

1  (( 

—  T52 

it 

.0327234632 

(( 

«  56'  15", 

-ih  " 

ti 

*  .0163622792 

(I 

"  28'  7"  30'", 

1  (( 

u 

.0081812080 

a 

"  14'  3"  45"', 

1   u 

a 

.0040906112 

u 

«  V  1"  521"', 

1  (( 

=  37T72 

a 

.0020453068 

etc.  etc. 

It  is  obvious  that  an  arc  so  small  as  seven  minutes  of  a 
degree  can  differ  but  very  little  from  its  chord ;  therefore, 
if  we  take  .002045307  to  be  the  true  value  of  the  ,^3  of 
!.he  circumference,  the  whole  circumference  must  be  the 


BOOK    V.  141 

product  of  .002045307  by  3072,  which  is  6.283183104  - 
circumference  whose  radius  is  unity.  The  half  of  this, 
3.141592552,  is  the  semi-circumference,  the  more  exact 
value  of  which,  as  stated,  (Prop.  6),  is  3.141592653. 

The  value  of  the  half  circumference  being  now  deter- 
mined, if  that  of  any  arc  whatever  be  required,  we  have 
merely  to  divide  3.141592,  etc.,  by  10800,  the  number  of 
minutes  in  a  semi-circumference,  and  multiply  the  quo- 
tient by  the  number  of  minutes  in  the  arc  whose  length 
is  required. 

But  this  investigation  has  been  carried  far  enough  for 
our  present  purposes.  It  will  be  resumed  under  the 
subject  of  Trigonometry. 

We  insert  the  following  beautiful  theorem  for  the  tri- 
section  of  an  arc,  although  not  necessary  for  practical 
application.  Those  not  acquainted  with  cubic  equations 
may  omit  it. 

PROPOSITION   VIII. —  THEOREM. 

Given,  the  chord  of  any  arc,  to  determine  the  chord  of  one 
third  of  such  arc. 

Let  AE  be  the  given  chord,  and 
conceive  its  arc  divided  into  three 
equal  parts,  as  represented  by  AB, 
BD,  and  BE. 

Through  the  center  draw  BCCr,  and 
draw  AB.  The  two  A's,  CAB  and 
ABF,  are  equiangular;  for,  the  angle 
FAB,  being  at  the  circumference,  is 
measured  by  one  half  the  arc  BE,  which  is  equal  to  AB, 
and  the  angle  BOA,  being  at  the  center,  is  measured  by 
the  arc  AB ;  therefore,  the  angle  FAB  =  the  angle  BOA ; 
but  the  angle  OB  A  or  FBA,  is  common  to  both  tri- 
angles ;  therefore,  the  third  angle,  CAB,  of  the  one  tri- 
angle, is  equal  to  the  third  angle,  AFB,  of  the  other, 


U2  GEOMETRY. 

(Th.  12,  B.  I,  Cor.  2),  and  the  two  triangles  are  equi- 
angular and  similar. 

But  the  A  ACB  is  isosceles;  therefore,  the  A  AFB  is 
alsc  isosceles,  and  AB  =  A F,  and  we  have  the  following 
proportions : 

CA  :  AB  : :  AB  :  BF. 

Now,  let  AF=  e,  AB  =  x,  AC=  1.  Then  AF=  x>  and 
EF—  c  —  x,  and  the  proportion  becomes, 

1  :  x  : :  x  :  BF,     Hence,  BF=  x\ 

Also,  Fa  =  2  —  x\ 

As  AE  and  i?6r  are  two  chords  intersecting  each  other 
at  the  point  F,  we  have, 

GFx  FB  =  AFx  FE,  (Th.  17,  B.  HI). 

That  is,  (2  —  x2)  x*  =  x  (c  —  x) ; 

or,  #3  —  3a;  =  —  <?. 

If  we  suppose  the  arc  AE  to  be  60  degrees,  then  c  =  1, 
and  the  equation  becomes  or5  —  3x  =  —  1 ;  a  cubic  equa- 
tion, easily  resolved  by  Horner's  method,  (Robinson's 
New  University  Algebra,  Art.  464),  giving  x  =  .347296  + 
the  chord  of  20°.  This  again  may  be  taken  for  the  value 
of  c,  and  a  second  solution  will  give  the  chord  of  6°  40', 
and  so  on,  trisecting  successively  as  many  times  as  we 
please. 

PRACTICAL   PROBLEMS. 

The  theorems  and  problems  with  which  we  have  been 
thus  far  occupied,  relate  to  plane  figures;  that  is,  to 
figures  all  of  whose  parts  are  situated  in  the  same  plane. 
It  yet  remains  for  us  to  investigate  the  intersections  and 
relative  positions  of  planes ;  the  relations  and  positions 
of  lines  with  reference  to  planes  in  which  they  are  not 
contained ;  and  the  measurements,  relations,  and  proper- 
ties of  solids,  or  volumes.  But  before  we  proceed  to  this, 
it  is  deemed  advisable  to  give  some  practical  problems 
for  the  purpose  of  exercising  the  powers  of  the  student, 


BOOK  V.  143 

and  of  fixing  in  his  mind  those  general  geometrical  prin- 
ciples with  which  we  must  now  suppose  him  to  be 
acquainted.    . 

1.  The  base  of  an  isosceles  triangle  is  6,  and  the  oppv>- 
site  angle  is  60° ;  required  the  length  of  each  of  the  other 
two  equal  sides,  and  the  number  of  degrees  in  each  of 
the  other  angles. 

2.  One  angle  of  a  right-angled  triangle  is  30° ;  what 
is  the  other  angle  ?  Also,  the  least  side  is  12,  what  is 
the  hypotenuse  ? 

A        J  The  hypotenuse  is  24,  the  double  of  the  least 
***'   \      side.     Why? 

3.  The  perpendicular  distance  between  two  parallel 
lines  is  10 ;  what  angles  must  a  line  of  20  make  with 
these  parallels  to  extend  exactly  from  the  one  to  the 
other  ?  Arts.  The  angles  must  be  30°  and  150°. 

4.  The  perpendicular  distance  between  two  parallels 
is  20  feet,  and  a  line  is  drawn  across  them  at  an  angle  of 
45°  ;  what  is  its  length  between  the  parallels  ? 

Ans.  20^2. 

5.  Two  parallels  are  8  feet  asunder,  and  from  a  point 
in  one  of  the  parallels  two  lines  are  drawn  to  meet  the 
other ;  the  length  of  one  of  these  lines  is  10  feet,  and 
that  of  the  other  15  feet ;  what  is  the  distance  between 
the  points  at  which  they  meet  the  other  parallel  ? 

Ans.  6.69  ft.,  or  18.69  ft.     (See  Th.  39,  B.  I). 

6.  Two  parallels  are  12  feet  asunder,  and,  from  a  point 
on  one  of  them,  two  lines,  the  one  20  feet  and  the  other 
18  feet  in  length,  are  drawn  to  the  other  parallel ;  what 
is  the  distance  between  the  two  lines  on  the  other  parallel, 
and  what  is  the  area  of  the  triangle  so  formed  ? 

c  The  distance  on  the  other  parallel  is  29.416 
Ans.    <      feet,  or  2.584  feet;  and  the  area  of  the  tri 
1      angle  is  176.496,  or  15.504  square  feet. 

7.  The  diameter  of  a  circle  is  12,  and  a  chord  of  the 


144  GEOMETRY. 

circle  is  4;   what  is  the  length   of  the  perpendicular 
drawn  from  the  center  to  this  chord  ?   (See  Th.  3,  B.  III). 

Ans.  4^2. 

8.  Two  parallel  chords  in  a  circle  were  measured  and 
found  to  be  8  feet  each,  and  their  distance  asunder  was 
6  feet ;  what  was  the  radius  of  the  circle  ? 

Ans.  5  feet. 

9.  Two  chords  on  opposite  sides  of  the  center  of  a 
cirsle  are  parallel,  and  one  of  them  has  a  length  of  It) 
and  the  other  of  12  feet,  the  distance  between  them 
being  14  feet.     What  is  the  diameter  of  the  circle  ? 

Ans.  20  feet. 

10.  An  isosceles  triangle  has  its  two  equal  sides,  15 
each,  and  its  base  10.  What  must  be  the  altitude  of  a 
right-angled  triangle  on  the  same  base,  and  having  an 
equal  area? 

11.  From  the  extremities  of  the  base  of  any  triangle, 
draw  lines  bisecting  the  other  sides ;  these  two  lines  in- 
tersecting within  the  triangle,  will  form  another  triangle 
on  the  same  base.  How  will  the  area  of  this  new  tri- 
angle compare  with  that  of  the  whole  triangle  ? 

Ans.  Their  areas  will  be  as  3  to  1. 

12.  Two  parallel  chords  on  the  same  side  of  the  center 
of  a  circle,  whose  diameter  is  32,  are  measured  and  found 
to  be,  the  one  20,  and  the  other  8.  How  far  are  they 
asunder?  Ans.  ^240"—  ^156"=  3  +. 

If  we  suppose  the  two  chords  to  be  on  opposite  sides  of  the 
center,  their  distance  apart  will  then  be  v'240  -j-  v/156  =  ]  5.49  -f- 
12.49  =  27.98. 

13.  The  longer  of  the  two  parallel  sides  of  a  trapezoid 
is  12,  the  shorter  8,  and  their  distance  asunder  5.  What 
is  the  area  of  the  trapezoid  ?  and  if  we  produce  the  two 
inclined  sides  until  they  meet,  what  will  be  the  area  of 
the  triangle  so  formed  ? 

Ans.  Area  of  trapezoid,  50 ;  area  of  triangle,  40 ;  area 
of  triangle  and  trapezoid,  90. 


BOOK    V. 


145 


14.  The  base  of  a  triangle  is  697,  one  of  the  sides  is 
534,  and  the  other  813.  If  a  line  be  drawn  bisecting  the 
angle  opposite  the  base,  into  what  two  parts  will  the 
bisecting  line  divide  the  base  ?     (See  Th.  24,  B.  II). 

A        (  The  greater  part  will  be  420.684 ; 
m'   \  The  less  "  "     276.316. 

15.  Draw  three  horizontal  parallels,  making  the  dis- 
tance between  the  two  upper  parallels  7,  and  that  be- 
tween the  middle  and  lower  parallels  9 ;  then  place  be- 
tween the  upper  parallels  a  line  equal  to  10,  and  from 
the  point  in  which  it  meets  the  middle  parallel  draw  to 
the  lower  a  line  equal  to  11,  and  join  the  point  in  which 
this  last  line  meets  the  lower  parallel,  with  the  point  in 
the  upper  parallel,  from  which  the  line  10  was  drawn. 
Required  the  length  of  this  line,  and  the  area  of  the 
triangle  formed  by  it  and  the  two  lines  10  and  11. 

The  adjoining  figure 
will  illustrate.  Let  A  be 
the  point  on  the  upper 
parallel  from  which  the 
line  10  is  drawn.  Then, 
AF  =  7,  AB  =  10, 
FB  =  \/l00  — 19  = 


•51. 

BI1  =  FD  =  9,  BC 
=  U,HC=  </l21_81 
=  %/40. 

Whence,  DC  =  ^51 
f  ^40. 

AC2  -  (V5l  4.  </40)8  -f  (16)a;  AG  =  20.89,  Ans. 

The  area  of  the  triangle,  ABC,  can  be  determined  by  first  find- 
ing the  area  of  the  trapezoid,  ABHD,  then  the  area  of  the  trian- 
gle, BHC,  and  from  their  sum  subtracting  the  area  of  the  triangle, 
ADC. 

16.  Construct  a  triangle  on  a  base  of  400,  one  of  the 
angles  at  the  base  being  80°,  and  the  other  70° ;  and 
13  k 


Ans. 


146  GEOMETRY. 

determine  the  third  angle,  and  the  area  of  the  triangle 
thus  constructed. 

rThe  third  angle  is  30°,  and  as  nearly  as  our 
scale  of  equal  parts  can  determine  for  us,  the 
side  opposite  the  angle  80°  is  787,  and  that 
opposite  70°  is  740. 
The  exact  solution  of  problems  like  the  last,  except  in  a  few  par- 
ticular cases,  requires  a  knowledge  of  certain  lines  depending  on 
the  angles  of  the  triangle.  The  properties  and  values  of  these  lines 
are  investigated  in  trigonometry ;  and  as  we  are  not  yet  supposed 
to  be  acquainted  with  them,  we  must  be  content  with  the  approxi- 
mate solutions  obtained  by  the  constructions  and  measurements 
made  with  the  plane  scale. 

17.  If  we  call  the  mean  radius  of  the  earth  1,  the 
mean  distance  of  the  moon  will  he  60 ;  and  as  the  mean 
distance  of  the  sun  is  400  times  the  distance  of  the 
moon,  its  distance  will  he  400  times  60.  The  sun  and 
moon  appear  to  have  the  same  diameter;  supposing, 
then,  the  real  diameter  of  the  moon  to  be  2160  miles, 
what  must  he  that  of  the  sun  ? 


Let  E  be  the  center  of  the  earth,  M  that  of  the  moon,  and  S 
that  of  the  sun,  and  suppose  ENP  to  be  a  line  from  the  -center  of 
the  earth,  touching  the  moon  and  the  sun. 

Then,  EM  :  MN  : :  ES  :  SP-, 

but  MNia  the  radius  of  the  moon,  and  SP  that  of  the  sun.     Mul- 
tiplying the  consequents  by  2,  the  above  proportion  becomes 
EM:  2MN  ::  ES  :  2SP-, 

or  in  numbers,      60  :  2160   ::  400  X  60  :  2SP; 
whence,  2  JSP  =  sun's  diameter  =  864000  miles,  Ans. 
18.  In  Problem  15,  suppose  BO  to  be  drawn  on  the 
other  side  of  BR,  what,  then,  will  be  the  value  of  A  O, 
and  what  the  area  of  the  triangle  AQB1 
Am     <AC=  16,021; 

1  A.rea  of  triangle,  £(9t/51  +  7V40). 


BOOK    V.  147 

19.  A  man  standing  40  feet  from  a  building  which  was 
24  feet  wide,  observed  that  when  he  closed  one  eye,  the 
width  of  the  building  just  eclipsed  or  hid  from  view  90 
rods  of  fence  which  was  parallel  to  the  width  of  the 
building;  what  was  the  distance  from  the  eye  of  the 
observer  to  the  fence  ?  Ans.  2475  feet. 

20.  Taking  the  same  data  as  in  the  last  problem,  ex- 
cept that  we  will  now  suppose  the  direction  of  the  fence 
to  be  inclined  at  an  angle  of  45°  to  the  side  of  the 
building  which  we  see ;  what,  in  this  case,  must  be  the 
distance  between  the  eye  of  the  observer  and  the  remoter 
point  of  the  fence  ? 


Let  EF  be  the  width  of  the  house,  E  the  position  of  the  eye,  and 
AB  that  of  the  fence.  Draw  BD  perpendicular  to  EA  produced ; 
then,  since  the  triangle  ABE  is  right-angled  and  isosceles,  we  have 
AE  =  EB}  and  2AE*  =  AB2  =  (90)2;  BE  =  63.64  rods,  and  the 
similar  triangles  EFH  and  EEB  give  the  proportion 
EF  :  EF  : :  BE  :  EE  =  1750.1  feet; 
and  from  this  we  find 

EB2  mm  EE2  +  BE2  =  (63.64  x  323)2  -f  (1750.1)* 
Whence  EB  =  2040.94  -f  Ans. 

21.  In  a  right-angled  triangle,  ABO,  we  have  AB  m 
493,  AC=  1425,  and  BC  =  1338 ;  it  is  required  to  divide 
this  triangle  into  parts  by  a  line  parallel  to  AB,  whose 
areas  are  to  each  other  as  1  is  to  3.  How  will  the  sides 
AC  and  BO  be  divided  by  this  line  ?   (See  Th.  20,  B.  II). 

Ans.  Into  equal  parts. 

22.  In  a  right-angled  triangle,  ABO,  right-angled  at 
B,  the  base  AB  is  320,  and  the  angle  A  is  60° ;  required 
the  remaining  angle  and  the  other  sides. 

A       /The  angle  (7=30°; 
n8'   \  AC=  640;  BO=  554.24. 


148  GEOMETRY. 

23.  A  hunter,  wishing  to  determine  his  distance  from 
a  village  in  sight,  took  a  point  and  from  it  laid  off  two 
lines  in  the  direction  of  two  steeples,  wnich  he  supposed 
equally  distant  from  him,  and  which  he  knew  to  be  100 
rods  asunder.  At  the  distance  of  50  feet  on  each  line 
from  the  common  point,  he  measured  the  distance  be- 
tween the  lines,  and  found  it  to  be  5  feet  8  inches.  How 
far  was  he  from  the  steeples  ? 

5  ft.  8  in. :  100  rods : :  50  ft. :  distance.  c  14,55y  feet, 

or,  68  :  100  x  ^  x  12 : :  50 :  distance.        An8' 1      °r  nf arl? 
z  13  miles. 

24.  A  person  is  in  front  of  a  building  which  he  knowa 
to  be  160  feet  long,  and  he  finds  that  it  covers  10  minutes 
of  a  degree ;  that  is,  he  finds  that  the  two  lines  drawn 
from  his  eye  to  the  extremities  of  the  building  include 
an  angle  of  10  minutes.  What  is  his  distance  from  the 
building?  ^^   (      55;004  feet,  or 

'  (  more  than  10  miles. 
Remark. — The  questions  of  distance,  with  which  we  are  at  present 
occupied,  depend  for  their  solution  on  the  properties  of  similar  tri- 
angles. In  the  preceding  example  we  apparently  have  but  one  tri- 
angle, but  we  have  in  fact  two ;  the  second  being  formed  by  the  dis- 
tances unity  on  the  lines  drawn  from  the  eye  of  the  observer,  and  the 
line  which  connects  the  extremities  of  these  units  of  distance.  This 
last  line  may  be  regarded  as  the  chord  of  the  arc  10  minutes  to  the 
radius  unity.  We  have  seen  that  the  length  of  the  arc  180°  to  the 
radius  1,  is  3.1415926 ;  hence  the  chord  of  1°  or  60'  is  0.017453,  and 
of  10/  it  must  be  0.0029089.     Therefore,  by  similar  triangles,  we  have 

0.0029089  :  160  : :  1  :  Ans.  =  ^~. 

25.  In  the  triangle,  ABO,  we  have  given  the  angles 
A  =  32°,  and  B  =  84°.  The  side  AB  is  produced,  and 
the  exterior  angle  CBD  thus  formed,  is  bisected  by  the 
line  BE,  and  the  angle  A  is  also  bisected  by  the  line  AE, 
BE  and  AE  meeting  in  the  point  E.  "What  is  the  angle 
ft  and  what  is  the  relation  between  the  angles  O  and  E ! 

Ans.   (7=64°;  E=\  ft 


BOOK  V.  149 

26.  Suppose  a  line  to  be  drawn  in  any  direction  be- 
tween two  parallels.  Bisect  the  two  interior  angles  thus 
formed  on  either  side  of  the  connecting  line,  and  prove 
that  the  bisecting  lines  meet  each  other  at  right  angles, 
and  that  they  are  the  sides  of  a  right-angled  triangle  of 
which  the  line  connecting  the  parallels  is  the  hypotenuse. 

27.  If  the  two  diagonals  of  a  trapezoid  be  drawn, 
show  that  two  similar  triangles  will  be  formed,  the 
parallel  sides  of  the  trapezoid  being  homologous  sides 
of  the  triangles.  "What  will  be  the  relative  areas  of 
these  triangles  ? 

{The  triangles  will  be  to  each  other 
as  the  squares  on  the  parallel  sides 
of  the  trapezoid. 

28.  If  from  the  extremities  of  the  base  of  any  triangle, 
lines  be  drawn  to  any  point  within  the  triangle,  forming 
with  the  base  another  triangle ;  how  will  the  vertical 
angle  in  this  last  triangle  compare  with  that  in  the 
original  triangle  ? 

r  It  will  be  as  much  greater  than  the  angle 
in  the  original  triangle  as  the  sum  of 
angles  at  the  base  of  the  new  triangle  is 
less  than  the  sum  of  those  at  the  base 
of  the  first. 

29.  The  two  parallel  sides  of  a  trapezoid  are  12  and 
20,  respectively,  and  their  perpendicular  distance  is  8. 
If  a  line  whose  length  is  14.5  be  drawn  between  the  in- 
clined sides  and  parallel  to  the  parallel  sides,  what  is  the 
area  of  the  trapezoid,  and  what  the  area  of  each  part, 
respectively,  into  which  the  trapezoid  is  divided  ? 

Area  of  the  whole,  128  square  units; 
"     smaller  part,    33J  " 

I      "     larger        "       94J  " 

Dividing  line  at  the  distance  of  2J  from 
shortei  parallel  side. 
•80.   If  we  assume  the  diameter  of  the  earth  to  be 
13* 


150  *         GEOMETRY. 

7956  miles,  and  the  eye  of  an  observer  be  40  feet  above 
the  level  of  the  sea,  how  far  distant  will  an  object  be, 
that  is  just  visible  on  the  earth's  surface.  (Employ  Th* 
18,  B.  Ill,  after  reducing  miles  to  feet.) 

Ans.  40992  feet  =  7  miles  4032  feet. 

31.  The  diameter  of  a  circle  is  4 ;  what  is  the  area  of 
the  inscribed  equilateral  triangle  ?  Ans.  3^3. 

32.  Three  brothers,  whose  residences  are  at  the  ver- 
tices of  a  triangular  area,  the  sides  of  which  are  severally 
10,  11,  and  12  chains,  wish  to  dig  a  well  which  shall  be 
at  the  same  distance  from  the  residence  of  each.  Deter- 
mine the  point  for  the  well,  and  its  distance  from  their 
residences. 

Remark.  —  Construct  a  triangle,  the  sides  of  which  are,  respectively, 
10,  11,  and  12.  The  sides  of  this  triangle  will  be  the  chords  of  a  cir- 
cle whose  radius  is  the  required  distance.  To  find  the  center  of  this 
circle,  bisect  either  two  of  the  sides  of  the  triangle  by  perpendiculars, 
and  their  intersection  will  be  the  center  of  the  circle,  and  the  location 
of  the  well. 

Ans.  The  well  is  distant  6.405  chains,  nearly,  from  each 
residence. 

33.  The  base  of  an  isosceles  triangle  is  12,  and  the 
equal  sides  are  20  each.  "What  is  the  length  of  the  per- 
pendicular from  the  vertex  to  the  base;  and  what  the 
area  of  the  triangle  ? 

Ans.  Perpendicular,  19.07;  area,  (19.07)  x  6. 

34.  The  hypotenuse  of  a 
right-angled  triangle  is  45 
inches,  and  the  difference  be- 
tween the  two  sides  is  8.45 
inches.  Construct  the  triangle. 

Suppose  the  triangle  drawn  and 
^presented  by  ABC,  DC  being  the 
difference  between  the  two  sides. 

Now,  by  inspection,  we  discover  the 
steps  to  be  taken  for  the  construc- 
tion of  the  triangle    As  AD  =  AB, 


BOOK    V.  151 

the  angle  ADB,  must  be  equal  to  the  angle   DBA,  and  each  equal 
to  45°. 

Therefore,  draw  any  line,  AC,  and  from  an  assumed  point  in  it 
as  D,  draw  BD,  making  the  angle  ADB  =  45°.  Take  from  a 
scale  of  equal  parts,  8.45  inches,  and  lay  them  off  from  D  to  C,  and 
with  C  as  a  center,  and  CB  =  45  inches  as  a  radius,  describe  an 
arc  cutting  BD  in  B.  Draw  CB,  and  from  B,  draw  BA  at  right 
angles  to  A C)  then  is  ABC the  triangle  sought. 

Ans.  AB  =27.3;  AC=  35.76,  when  carefully  constructed. 

35.  Taking  the  same  triangle  as  in  the  last  problem,  if 
we  draw  a  line  bisecting  the  right  angle,  where  will  it 
meet  the  hypotenuse? 

Ans.  19.5  from  B ;  and  25.5  from  Q. 

36.  The  diameters  of  the  hind  and  fore  wheels  of  a 
carriage,  are  5  and  4  feet,  respectively ;  and  their  centers 
are  6  feet  asunder.  At  what  distance  from  the  fore  wheels 
will  the  line,  passing  through  their  centers,  meet  the 
ground,  which  is  supposed  level  ?  Ans.  24  feet. 

37.  If  the  hypotenuse  of  a  right-angled  triangle  is  35, 
and  the  side  of  its  inscribed  square  12,  what  are  its  sides? 

Ans.  28  and  21. 

38.  What  are  the  sides  of  a  right-angled  triangle 
having  the  least  hypotenuse,  in  which  if  a  square  be  in- 
scribed, its  side  will  be  12  ? 

r  The  sides  are  equal  to  24  each,  and  the 
Ans.   <      least  hypotenuse  is  double  the  diagonal 
I     of  the  square. 

39.  The  radius  6f  a  circle  is  25 ;  what  is  the  area  cf  a 
sector  of  50°  ? 

Remark.  —  First  find  the  length  of  an  arc  of  50°  in  a  circle  Vncsa 

radius  is  unity.     Then  25  times  that  will  be  the  length  of  an  arc  of 

the  same  number  of  degrees  in  a  circle  of  which  the  radius  is  25. 

,     ,        -.o      ,-          •          3.14159265 
Length  of  arc  1°  radius  unity  = ^r — . 

(i  «  goo       u        «     _  1.04/1J755      g 

6 

•Area  of  sector  =  L04719755  X  125  x  f  =  272  7077.  Ans. 
o  2 


152  GEOMETRY. 


BOOK   VI. 


ON  THE  INTERSECTIONS  OF  PLANES,  AND  THE  EEL. 
ATIVE  POSITIONS  OF  PLANES  AND  OF  PLANES 
AND  LINES. 

DEFINITIONS. 

A  Plane  has  been  already  defined  to  be  a  surface,  such 
that  the  straight  line  which  joins  any  two  of  its  points 
will  lie  entirely  in  that  surface.     (Def.  9,  page  9.) 

1.  The  Intersection  or  Common  Section  of  two  planes  is 
the  line  in  which  they  meet. 

2.  A  Perpendicular  to  a  Plane  is  a  line  which  makes 
right  angles  with  every  line  drawn  in  the  plane  through 
the  point  in  which  the  perpendicular  meets  it;  and,  con- 
versely, the  plane  is  perpendicular  to  the  line.  The 
point  in  which  the  perpendicular  meets  the  plane  is 
called  the  foot  of  the  perpendicular. 

3.  A  Diedral  Angle  is  the  separation  or  divergence  of 
two  planes  proceeding  from  a  common  line,  and  is  meas- 
ured by  the  angle  included  between  two  lines  drawn 
one  in  each  plane,  perpendicular  to  their  common  sec- 
tion at  the  same  point. 

The  common  section  of  the  two  planes  is  called  the 
edge  of  the  angle,  and  the  planes  are  its  faces, 

4.  Two  Planes  are  perpendicular  to  each  other,  when  their 
diedral  angle  is  a  right  angle. 

5.  A  Straight  Line  is  parallel  to  a  plane,  when  it  will 
not  meet  the  plane,  however  far  produced. 


BOOK  VI.  153 

6.  Two  Planes  are  parallel,  when  they  will  not  intersect, 
however  far  produced  in  all  directions. 

7.  A  Solid  or  Polyedral  Angle  is  the  separation  or  diver- 
gence of  three  or  more  plane  angles,  proceeding  from  a 
common  point,  the  two  sides  of  each  of  the  plane  angles 
being  the  edges  of  diedral  angles  formed  by  these  plane 
angles. 

The  common  point  from  which  the  plane  angles  pro- 
ceed h  called  the  vertex  of  the  solid  angle,  and  the  inter- 
sections of  its  bounding  planes  are  called  its  edges, 

8.  A  Triedral  Angle  is  a  solid  angle  formed  by  three 
plane  angles. 

THEOREM   I. 

Two  straight  lines  which  intersect  each  other,  two  parallel 
straight  lines,  and  three  points  not  in  the  same  straight  line, 
will  severally  determine  the  position  of  a  plane. 

Let  AB  and  A  0  be  two  lines 
Intersecting  each  other  at  the 
point  A ;  then  will  these  lines 
determine  a  plane.  For,  conceive 
a  plane  to  be  passed  through  AB, 
and  turned  about  AB  as  an  axis 
until  it  contains  the  point  0  in  the  line  AC  The  plane, 
in  this  position,  contains  the  lines  AB  and  AC,  and  will 
contain  them  in  no  other.  Again,  let  AB  and  BE  be 
two  parallel  straight  lines,  and  take  at  pleasure  two 
points,  A  and  B,  in  the  one,  and  two  points,  D  and  E, 
in  the  other,  and  draw  AE  and  BD.  The  last  lines,  AB,  AE, 
or  the  lines  AB,  DB  from  what  precedes,  determine  the  posi- 
tion of  the  parallels  AB,  BE.  And  again,  if  A,  B,  and  0 
be  three  points  not  in  the  same  straight  line,  and  we  draw 
the  lines  AB  and  AC,  it  follows,  from  the  first  part  of  this 
proposition,  that  these  points  fix  the  plane. 


154 


GEOMETRY. 


Cor.  A  straight  line  and  a  point  out  of  ic  determine 
the  position  of  a  plane. 

THEOREM   II. 

If  two  planes  meet  each  other,  their  common  points  will  be 
found  in,  and  form  one  straight  line. 

Let  B  and  D  he  any  two  of  the 
points  common  to  the  two  planes, 
and  join  these  points  by  the  straight 
line  BB ;  then  will  BB  contain  all 
the  points  common  to  the  two  planes, 
and  he  their  intersection.  For,  suppose  the  planes  have 
a  common  point  out  of  the  line  BB;  then,  (Cor.  Th.  1), 
since  a  straight  line  and  a  point  out  of  it  determine  a 
plane,  there  would  be  two  planes  determined  by  this  one 
line  and  single  point  out  of  it,  which  is  absurd.  Hence 
the  common  section  of  two  planes  is  a  straight  line. 

Remark. — The  truth  of  this  proposition  is  implicitly  assumed  in  the 
definitions  of  this  Book. 


THEOREM    III. 

If  a  straight  line  stand  at  right  angles  to  each  of  two  other 
straight  lines  at  their  point  of  intersection,  it  will  be  at  right 
angles  to  the  plane  of  those  lines. 

Let  AB  stand  at  right  angles  to  i^Fand 
CB,  at  their  point  of  intersection  A.  Then 
AB  will  be  at  right  angles  to  any  other 
line  drawn  through  A  in  the  plane,  pass- 
ing through  EF,  OB,  and,  of  course,  at 
right  angles  to  the  plane  itself.     (Def.  2.) 

Through  A,  draw  any  line,  A  G,  in  the 
plane  EF,  CB,  and  from  any  point  G,  draw  GH  parallel 
to  AB.    Take  HF  =  AH,  and  join  F  and  G  and  produce 
FG  to  B.    Because  HG  is  parallel  to  AB,  we  have 
FH  :  HA  : :  FG  :  GB. 


BOOK  VI.  155 

But,  in  this  proportion,  the  first  couplet  is  a  ratic  of 
equality;  therefore  the  last  couplet  is  also  a  ratio  of 
equality, 

That  is,  FG  =  GD,  or  the  line  FD  is  bisected  in  G. 

Draw  BD,  BG,  and  BF. 

"Now,  in  the  triangle  AFD,  as  the  base  FD  is  bisected 
in  G,  we  have, 

AF2  +~AD2  -  2AG2  +  2GF2    (1)    (Th.  42,  B.  I). 

Also,  as  DF  is  the  base  of  the  a  BDF,  we  have  by  the 
same  theorem, 

"SF  +~BD2  =  2BG2  +  2GF%  (2) 

By  subtracting  ( 1 )  from  ( 2 ),  and  observing  that  BF2 — 
AF2  =  AB\  because  BAF  is  a  right  angle ;  and  BD2  — 
AD1  —  AB2,  because  BAB  is  a  right  angle,  we  shall  have, 


AB  +  AB  =  2BG1  —  2AG 


lhviding  by  2,  and  transposing  AG  ,  and  we  have, 


AB'  +  AG  =BG  . 

This  last  equation  shows  that  BAG  is  a  right  angle. 
But  AG  is  any  line  drawn  through  A,  in  the  plane  FF, 
CD ;  therefore  AB  is  at  right  angles  to  any  line  in  the 
plane,  and,  of  course,  at  right  angles  to  the  plane  itself 

Cor.  1.  The  perpendicular  BA  is  shorter  than  any  of 
the  oblique  lines  BF,  BG,  or  BD,  drawn  from  the  point 
B  to  the  plane ;  hence  it  is  the  shortest  distance  from  a 
point  to  a  plane. 

Cor.  2.  But  one  perpendicular  can  be  erected  to  a  plane 
from  a  given  point  in  the  plane ;  for,  if  there  could  be 
two,  the  plane  of  these  perpendiculars  would  intersect 
the  given  plane  in  some  line,  as  AG,  and  both  the  per- 
pendiculars would  be  at  right  angles  to  this  intersection 
at  the  same  point,  which  is  impossible. 

Cor.  3.  But  one  perpendicular  can  be  let  fall  from  a 
given  point  out  of  a  plane  on  the  plane ;  for,  if  there  can 


156 


GEOMETKY. 


be  two,  let  BCr  and  BA  be  such  perpendiculars,  then 
would  the  triangle  BAGf  be  right  angled  at  both  A  and 
<r,  which  is  impossible. 


THEOREM   IV. 

If  from  any  'point  of  a  perpendicular  to  a  plane,  oblique 
lines  be  drawn  to  different  points  in  the  plane,  those  oblique 
lines  which  meet  the  plane  at  equal  distances  from  the  foot  of 
the  perpendicular  are  equal;  and  those  which  meet  the  plane 
at  unequal  distances  from  the  foot  of  the  perpendicular  are 
unequal,  the  greater  distances  corresponding  to  the  longer 
oblique  lines, 

Take  any  point  B  in 
the  perpendicular  BA  to 
the  plane  ST,  and  draw 
the  oblique  lines  BO, 
BD,  and  BE,  the  points 
C,  J),  and  E,  being  equally 
distant  from  A,  the  foot 
of  the  perpendicular. 
Produce  AE  to  F,  and 
draw  BF;  then  will  BC=  BD  —  BE,  and  BF>  BE. 

For,  the  triangles  BAG,  BAB,  and  BAE  are  all  right- 
angled  at  J.,  the  side  BA  is  common,  and  AC=  AD=AE 
by  construction,  hence,  (Th.  16,  B.  I),  BC=BD  =  BE. 
Moreover,  since  AF>  AE,  the  oblique  line  BF>  BE. 

Cor.  If  any  number  of  equal  oblique  lines  be  drawn 
from  the  point  B  to  the  plane,  they  will  all  meet  the 
plane  in  the  circumference  of  a  circle  having  the  foot  of 
the  perpendicular  for  its  center.  It  follows  from  this, 
that,  if  three  points  be  taken  in  a  plane  equally  distant 
from  a  point  out  of  it,  the  center  of  the  circle  whose  cir* 
cumference  passes  through  these  points  will  be  the  foot  of 
*he  ]aerx>endicular  drawn  from  the  point  to  the  plane. 


BOOK    VI.  157 


THEOREM   V. 


The  line  which  joins  any  point  of  a  perpendicular  to  a 
plane,  with  the  point  in  which  a  line  in  the  plane  is  inter- 
sected, at  right  angles,  by  a  line  through  the  foot  of  the  per- 
pendicular, will  be  at  right  angles  to  the  line  in  the  plane 

Let  AB  be  perpendic- 
ular to  the  plane  ST,  and 
AD  a  line  through  its  foot 
at  right  angles  to  EF,  a  line 
in  the  plane.  Connect  D 
with  any  point,  as  B,  of  the 
perpendicular;  and  BD  will 
be  perpendicular  to  EF. 

Make  DF '=  DE,  and  join  B  to  the  points  E  and 
F.  Since  DE  =  DF,  and  the  angles  at  D  are  right 
angles,  the  oblique  lines,  AE  and  AF,  are  equal ;  and, 
since  AE=  AF,  we  have,  (Th.  4),  BE=BF;  therefore 
the  line  BD  has  two  points,  B  and  D,  each  equally  distant 
from  the  extremities  E  and  F  of  the  line  EF,  and  hence 
BD  is  perpendicular  to  EF  at  its  middle  point  D. 

Cor.  Since  FD  is  perpendicular  to  the  two  lines  AD 
und  BD  at  their  intersection,  it  is  perpendicular  to  their 
plane  ADB,  (Th.  3). 

ScnoLiuM.  —  The  inclination  of  a  line  to  a  plane  is  measured  by  the 
angle  included  between  the  given  line  and  the  line  which  joins  the 
point  in  which  it  meets  the  plane  and  the  foot  of  the  perpendicular 
drawn  from  any  point  of  the  line  to  the  plane ;  thus,  the  angle  BFA  ia 
the  inclination  of  the  line  BF  to  the  plane  ST. 

THEOREM   VI. 

If  either  of  two  parallels  is  perpendicular  to  a  plane,  the 
other  is  also  perpendicular  to  the  plane. 

Let  BA  and  ED  be  two  parallels,  of  which  one,  BA, 
is  perpendicular  to  the  plane  ST;  then  will  the  other  also 
be.  perpendicular  to  the  same  plane. 
14 


158 


GEOMETRY. 


The  two  parallels  de- 
termine a  plane  which 
intersects  the  given  plane 
in  AD;  through  D  draw 
MN  perpendicular  to 
AD;  then,  (Cor.,  Th.  5,) 
will  MN  be  perpendicu- 
lar to  the  plane  BAD, 
and  the  angle  MDE  is 

therefore  a  right  angle ;  but  ED  A  is  also  a  right  angle, 
since  BA  and  ED  are  parallel,  and  BAD  is  a  right  angle 
by  hypothesis;  hence,  ED  is  perpendicular  to  the  two 
lines  MD  and  AD  in  the  plane  ST;  it  is  therefore  perpen- 
dicular to  the  plane,  (Th.  3). 

Cor.  1.  The  converse  of  this  proposition  is  also  true , 
that  is,  if  two  straight  lines  are  both  perpendicular  to  the  same 
plane,  the  lines  are  parallel. 

For,  suppose  BA  and  ED  to  be  two  perpendiculars ;  if 
not  parallel,  draw  through  D  a  parallel  to  BA,  and  this 
last  line  will  be  perpendicular  to  the  plane ;  but  ED  is 
a  perpendicular  by  hypothesis,  and  we  should  have  two 
perpendiculars  erected  to  the  plane  at  the  same  point, 
which  is  impossible,  (Cor.  2,  Th.  3). 

Cor.  2.  If  two  lines  lying  in  the  same  plane  are  each 
parallel  to  a  third  line  not  in  the  same  plane,  the  two 
lines  are  parallel.  For,  pass  a  plane  perpendicular  to 
the  third  line,  and  it  will  be  perpendicular  to  each  of  tlva 
others ;  hence  they  are  parallel , 


THEOREM    VII, 


A  straight  line  is  parallel  to  a  plane,  when  it  is  parallel 
to  a  line  in  the  plane. 

Suppose  the  line  MX  to  be  parallel  to  the  line  CD,  in 
the  plane  ST;  then  will  Jf^be  parallel  to  the  plane  ST 


BOOK    VI. 


159 


For,  OB  being   in  the  plane 

ST,   and    at    the    same    time     g 

parallel  to  MN,  it  must  be  the 

intersection   of  the   pl°.ne   of 

these  parallels  with  the  plane 

ST;   hence,  if  MN  meet  the 

plane  ST,  it  must  do  so  in  the  T 

line  OB,  or  OB  produced;  but  MN  and  OB  are  parallel, 

and  cannot  meet;  therefore  MN,  nowever  far  produced, 

can  have  no  point  in  the  plane  ST,  and  hence,  (Def.  5),  it 

is  parallel  to  this  plane. 

THEOREM    VIII. 

If  two  lines  are  parallel,  they  will  be  equally  inclined  to 
any  given  plane. 

Let  AB  and  OB  be 
two  parallels,  and  ST 
any  plane  met  by  them 
in  the  points  A  and 
0;  then  will  the  lines 
AB  and  OB  be  equally 
inclined  to  the  plane 
ST. 

For,  take  any  distance,  AB,  on  one  of  these  parallels, 
and  make  OB  =  AB,  and  draw  A  0  and  BB.  From  the 
points  B  and  B  let  fall  the  perpendiculars,  BE  and  BF, 
on  the  plane ;  join  their  feet  by  the  line  BF,  and  draw 
AB  and  OF. 

Now,  since  AB  is  equal  and  parallel  to  OB,  ABBOh 
ft  parallelogram,  and  BB  is  equal  and  parallel  to  A  0, 
and  BB  is  parallel  to  the  plane  ST,  (Th.  7) ;  and,  since 
BE  and  BF  are  both  perpendicular  to  this  plane,  they 
are  parallel ;  but  BB  and  EF  are  in  the  plane  of  these 
parallels;  and  as  EF  is  in  the  plane  ST,  and  BB  is 
parallel  to  this  plane,  these  two  lines  must  be  parallel 
and  equal,  and  BBFE  is  also  a  parallelogram      Now, 


160  GEOMETRY. 

we  have  shown  that  BD  is  equal  and  parallel  to  AC,  and 
EF  equal  and  parallel  to  BD;  hence,  (Cor.  2,  Th.  6), 
EFis  equal  and  parallel  to  AC,  and  ACFE  is  a  parallel- 
ogram, and  AE  —  CF.  The  triangles  J.^J/  and  CDF 
have,  then,  the  sides  of  the  one  equal  to  the  sides  of  the 
other,  each  to  each,  and  their  angles  are  consequently 
equal;  that  is,  the  angle  BAE  is  equal  to  the  angle 
DCF;  but  these  angles  measure  the  inclination  of  the 
lines  AB  and  CD  to  the  plane  ST,  (Scholium,  Th.  5). 

Scholium.  —  The  converse  of  this  proposition  is  not  generally  true  ; 
that  is,  straight  lines  equally  inclined  to  the  same  plane  are  not  neces- 
sarily parallel. 

THEOREM  IX. 

The  intersections  of  two  parallel  planes  by  a  third  plane, 
are  parallel. 

Let  the  planes  QR  and  ST  he  intersected  by  the  third 
plane,  AD :  then  will  the  intersections,  AB  and  CD,  be 
parallel. 

Since  the  lines  AB  and  CD  are  in  the  same  plane,  if 
they  are  not  parallel,  they  will 
meet  if  sufficiently  produced; 
but  they  cannot  meet  out  of  the 
planes  QR  and  ST,  in  which 
they  are  respectively  found; 
therefore,  any  point  common  to 
the  lines,  must  be  at  the  same 
time  common  to  the  planes ;  and 
since   the   planes    are    parallel, 

(hey  have  no  common  point,  and  the  lines,  therefore,  do 
not  intersect ;  hence  they  are  parallel. 

THEOREM    X. 

If  two  planes  are  perpendicular  to  the  same  straight  line, 
they  are  parallel  to  each  other. 

Let  QR  and  ST  be  two  planes,  perpendicular  to  the 
line  AB;  then  will  these  planes  be  parallel. 


'- 

— \ 

s 

] 

1— 

i" 

n 

"~"~ — ■ . 

\ 

BOOK  VI.  161 

For,  if  not  parallel,  suppose  M  to  be  a  point  in  their 
line  of  intersection,  and 
from  this  point  draw 
lines  to  the  extremities 
of  the  perpendicular  m" 
AB,  thus  forming  a  tri- 
angle, MAB.  Now, 
since  the  line  AB  is 
perpendicular   to    both 

planes,  it  is  perpendicular  to  each  of  the  lines  MA  and 
MB,  drawn  through  its  feet  in  the  planes,  (Def.  2); 
hence,  the  triangle  has  two  right  angles,  which  is  impos- 
sible ;  the  planes  cannot  therefore  meet  in  any  point  as 
M,  and  are  consequently  parallel. 

Cor.  Conversely :  The  straight  line  which  is  perpendicu- 
lar to  one  of  two  parallel  planes,  is  also  perpendicular  to  the 
other.  For,  if  AB  be  perpendicular  to  the  plane  QB, 
draw  in  the  other  plane,  through  the  point  in  which  the 
perpendicular  meets  it,  any  line,  as  AC.  The  plane  of 
the  lines  AB  and  A  C  will  intersect  the  plane  QB  in  the 
line  BD  ;  and  since  the  planes  are  parallel  by  hypothesis, 
the  lines  AC  and  BD  must  be  parallel,  (Th.  9) ;  but  the 
angle  DBA  is  a  right  angle ;  hence,  BA  C  must  be  a  right 
angle,  and  the  line  BA  is  perpendicular  to  any  line  what; 
ever  drawn  in  the  plane  through  the  point  A ;  BA  is 
therefore  perpendicular  to  the  plane  ST. 


THEOREM   XI. 

If  two  straight  lines  be  drawn  in  any  direction  through 
parallel  planes,  the  planes  will  cut  the  lines  proportionally. 

Conceive  three  planes  to  be  parallel,  as  represented 
in  the  figure,  and  take  any  points,  A  and  B,  in  the  first 
and  third  planes,  and  draw  AB,  the  line  passing  through 
the  second  plane  at  E. 

14*  L 


162 


GEOMETRY. 


Also,  take  any  other  two  points,  as 
0  and  D,  in  the  first  and  third  planes, 
and  draw  OB,  the  line  passing  through 
the  second  plane  at  F. 

Join  the  two  lines  by  the  diagonal 

AB,  which  passes  through  the  second 
plane  at  G.     Draw  BB,  EG,  GF,  and 

AC.  We  are  now  to  prove  that, 

AE  :  EB  : :  OF  :  FB. 
For  the  sake  of  brevity,  put  AG=X,  and  GD=  Y. 
As  the  planes  are  parallel,  BD  is  parallel  to  EG  ;  from 
the   two  triangles  ABD   and  AEG,  we  have,    (Th.  17, 
B.II); 

A  E  :  EB  : :  X  :  Y. 

Also,  as  the  planes  are  parallel,  GF  is  parallel  to  A  0, 
and  we  have, 

OF  :  FD  : :  X  :   Y. 

By  comparing  the  proportions,  and   applying  Th.  6, 
B.  II,  we  have 

AE  :  EB  : :  OF  :  FB. 


THEOREM    XII. 

If  a  straight  line  is  perpendicular  to  a  plane,  all  planes 
passing  through  that  line  will  be  perpendicular  to  the  plane. 

Let  MNbe  a  plane,  and  AB  a  per- 
pendicular to  it.  Let  BO  be  any 
other  plane,  passing  through  AB ; 
this  plane  will  be  perpendicular  to 
MM 

Let  BB  be  the  common  intersec- 
tion  of  the  two  planes,  and  from 
the  point  B,  draw  in  MN  BE  at  right  angles  to  DB. 

Then,  as  AB  is  perpendicular  to  the  plane  MN,  it  is 
perpendicular  to  every  line  in  that  plane,  passing  through 


M 


D^ 


BOOK    VI. 


163 


B ;  (Def.  2,) ;  therefore,  ABE  is  a  right  angle.  But  the 
angle  ABE,  (Def.  3),  measures  the  inclination  of  the  two 
planes ;  therefore,  the  plane  CB  is  perpendicular  to  the 
plane  MN;  and  thus  we  can  show  that  any  other  plane, 
passing  through  AB,  will  be  perpendicular  to  MN. 
Hence  the  theorem. 


THEOREM   XIII. 


If  two  planes  are  perpendicular  to  each  other,  and  a  line 
be  drawn  in  one  of  them  perpendicular  to  their  common  in- 
tersection, it  will  be  perpendicular  to  the  other  plane. 

Let  the  two  planes,  QR  and  ST,  be  perpendicular  to 
each  other,  and  draw  in  QR  the  line  CD  at  right  angles 
to  their  common  intersection,  R  V;  then  will  this  line  be 
perpendicular  to  the  plane  ST. 

In  the  plane  ST  draw  ED,  perpen- 
dicular to  VR  at  the  point  D. 
Then,  since  the  planes  QR  and  ST 
are  perpendicular  to  each  other,  the 
angle  ODE  is  a  right  angle,  and 
CD  is  perpendicular  to  the  two 
lines,  ED  and  VR,  passing  through 

its  foot  in  the  plane  ST.     CD  is  therefore  perpendicular 
to  the  plane  ST,  (Th.  3). 

Cor.  Conversely:  if  we  erect  a  perpendicular  to  the 
plane  ST,  at  any  point,  D,  of  its  intersection  with  the 
plane  QR,  this  perpendicular  will  lie  in  the  plane  QR. 
For,  if  it  be  not  in  this  plane,  we  can  draw  in  the  plane 
the  line  CD,  at  right  angles  to  VR',  and,  from  what  has 
been  shown  above,  CD  is  perpendicular  to  the^lane  ST, 
and  we  should  thus  have  two  perpendiculars  erected  to 
the  plane,  ST,  at  the  same  point,  which  is  impossible, 
(Cor.  2,  Th.  8V 


164 


GEOMETRY. 


THEOREM    XIV. 

The  common  intersection  of  two  planes,  'both  of  which  are 
perpendicular  to  a  third  plane,  will  also  be  perpendicular  to 
the  third  plane. 

Let  MN  be  the  common 
intersection  of  the  two 
planes,  QR  and  VX,  both 
of  which  are  perpendicular 
to  the  plane  ST;  then  will 
MZ^be  perpendicular  to  the 
plane  ST  For,  if  we  erect 
a  perpendicular  to  the  plane 
ST,  at  the  point  M,  it  will 
lie  in  both  planes   at  the 

same  time,  (Cor.  Th.  13);  and  this  perpendicular  must 
therefore  be  their  intersection.     Hence  the  theorem. 


THEOREM   XV. 

Parallel  straight  lines  included  between  parallel  planes, 
are  equal. 

Let  AB  and  2) (7 be  two  parallel  lines, 
included  by  the  two  parallel  planes, 
QR  and  ST;  then  will  AB  =  BO. 

For,  the  plane  AC,  of  the  parallel 
lines,  intersects  the  planes,  QR  and  ST, 
in   the   parallel  lines,   AB  and   BO, 
(Th.  9) ;    hence  ABGD  is  a  parallelogram,  and  its  oppo- 
site sides,  AB  and  BO,  are  equal. 

Oot.  It  follows  from  this  proposition,  that  parallel  planes 
are  everywhere  equally  distant ;  for,  two  perpendiculars 
drawn  at  pleasure  between  the  two  planes  are  parallel 
lines,  (Cor.  1,  Th.  6),  and  hence  are  equal ;  but  these  per- 
pendiculars measure  the  distance  between  the  planes. 


BOOK    Vl 


165 


THEOREM   XVI. 

Two  planes  are  parallel  when  two  lines  not  parallel,  lying 
in  the  one,  are  respectively  parallel  to  two  lines  lying  in  the 
other. 

Let  QR  and  ST  be 
two  planes,  the  first 
containing  the  two 
lines  AB  and  CD 
which  intersect  each 
other  at  E,  and  the 
second  the  two  lines 
LM  and  NO,  respect- 
ively parallel  to  AB 
and  CD-,  then  will 
these  planes  be  par- 
allel. 

For,  if  the  two  planes 
are  not  parallel,  they  must  intersect  when  sufficiently 
produced;  and  their  common  section  lying  in  both  planes 
at  the  same  time,  would  be  a  line  of  the  plane  QR.  Now, 
the  lines  AB  and  CD  intersect  each  other  by  hypothesis ; 
hence  one  or  both  of  them  must  meet  the  common  sec- 
tion of  the  two  planes.  Suppose  AB  to  meet  this  com- 
mon section ;  then,  since  AB  and  LM  are  parallel,  they 
determine  a  plane,  and  AB  cannot  meet  the  plane  ST  in 
a  point  out  of  the  line  LM;  but  AB  and  LM  being  par- 
allel, have  no  common  point.  Hence,  neither  AB  nor 
CD  can  meet  the  common  section  of  the  two  planes ;  that 
»,  they  have  no  common  section,  and  are  therefore  par- 
allel. 

Cor.  Since  two  lines  which  intersect  each  other,  deter- 
mine a  plane,  it  follows  from  this  proposition,  that  the 
plane  of  two  intersecting  lines  is  parallel  to  the  plane  of  two 
other  intersecting  lines  respectively  parallel  to  the  first  lines 


166 


GEOMETRS*. 


THEOREM    XVII. 

When  two  intersecting  lines  are  respectively  parallel  to  two 
other  intersecting  lines  lying  in  a  different  plane,  the  angles 
formed  by  the  last  two  lines  will  be  equal  to  those  formed  by 
the  first  two,  each  to  each,  and  the  planes  of  the  angles  will  be 
parallel. 

Let  QB  be  the  plane 
of  the  two  lines  AB 
and  CD,  which  inter- 
sect each  other  at  the 
point  E,  and  ST  the 
plane  of  the  two  lines 
LM  and  NO,  respect- 
ively parallel  to  AB 
and  QB ;  then  will  the 
[__BED  =  \_MPO, 
and  L  EEC  -  L 
MPN,  etc.,  and  the 
planes  QB  and  ST 
will  be  parallel. 

That  the  plane  of  one  set  of  angles  is  parallel  to  that 
of  the  other,  follows  from  the  Corollary  to  Theorem  16  ; 
we  have  then  only  to  show  that  the  angles  are  equal, 
each  to  each. 

Take  any  points,  B  and  B,  on  the  lines  AB  and  CD, 
and  draw  BD.  Lay  off  PM,  equal  to  and  in  the  same 
direction  with  JEB,  and  PO,  equal  to  and  in  the  same 
direction  with  BD,  and  draw  MO.  Now,  since  the  planes 
QR  and  ST  are  parallel,  and  BD  is  equal  and  parallel  to 
PO,  BDOP  is  a  parallelogram,  and  DO  is  equal  and  par 
allel  to  EP.  For  the  same  reason,  BM  is  equal  and 
parallel  to  EP;  therefore,  BDOM is  a  parallelogram,  and 
MO  is  equal  and  parallel  to  BD.  Hence  the  A's,  EBD 
and  PMO,  have  the  sides  of  the  one  equal  to  the  sidea 
of  the  other,  each  to  each;  they  are  therefore  equal,  and 


BOOK  VI.  167 

the  I  MPO  —  the  [    BED.    In  the  same  manner  it  can 
be  proved  that  \_BEC  =  \_MPN,  etc. 

Cor.  1.  The  plane  of  the  parallels  AB  and  LM  is  in- 
tersected by  the  plane  of  the  parallels  CD  and  NO,  in  the 
line  EP.  Now,  EB  and  ED  are  the  intersections  of  these 
two  planes  with  the  plane  QR,  and  PM  and  PO  are  the 
intersections  of  the  same  planes  with  the  parallel  plane 
ST.  It  has  just  been  proved  that  the  |__  BED  =  [_MPO. 
Hence,  if  the  diedral  angle  formed  by  two  planes,  be  cut  by 
two  parallel  planes,  the  intersections  of  the  faces  of  the  diedral 
angle  with  one  of  these  planes  will  include  an  angle  equal 
to  that  included  by  the  intersections  of  the  faces  with  the  other 
plane. 

Cor.  2.  The  opposite  triangles  formed  by  joining  the  cor- 
responding extremities  of  three  equal  and  parallel  straight 
lines  lying  in  different  planes,  will  be  equal  and  the  plane*  of 
the  triangles  will  be  parallel. 

Let  EP,  BM,  and  DO,  be  three  equal  and  parallel 
straight  lines  lying  in  different  planes.  By  joining  their 
corresponding  extremities,  we  have  the  triangles  EBD 
and  PMO.  Now,  since  EP  and  BM  are  equal  and 
parallel,  EBMP  is  a  parallelogram,  and  EB  is  equal  and 
parallel  to  PM;  in  the  same  manner,  we  show  that  ED 
is  equal  and  parallel  to  PO,  and  BD  to  MO;  hence  the 
triangles  are  equal,  ha  ring  the  three  sides  of  the  one, 
respectively,  equal  to  the  three  sides  of  the  other. 
That  their  planes  are  parallel,  follows  from  Cor.,  Theo 
rem  16. 

THEOREM   XVIII. 

Any  one  of  the  three  plane  angles  bounding  a  triedral 
angle,  is  less  than  the  sum  of  the  other  two. 

Let  A  be  the  vertex  of  a  solid  angle,  bounded  by  the 
three  plane  angles,  BAG,  BAD,  and  DAC;  then  will  any 
one  of  these  three  angles  be  less  than  the  sum  of  the 


168  GEOMETRY. 

other  two.     To  establish  this  proposition,  we  have  only 
to  compare  the  greatest  of  the  three 
angles  with  the  sum  of  the  other 
two. 

Suppose,  then,  BAG  to  be  the 
greatest  angle,  and  draw  in  its  plane  B« 
the    line   AE,  making    the    angle 
CAE  equal  to  the  angle  CAD.     On  "iT 

AE,  take  any  point,  E,  and  through  it  draw  the  line  CEB. 
Take  AD,  equal  to  AE,  and  draw  BD  and  DC. 

Now,  the  two  triangles,  CAD  and  CAE,  having  two 
sides  and  the  included  angle  of  the  one  equal  to  the  two 
sides  and  included  angle  of  the  other,  each  to  each,  are 
equal,  and  CE  =  CD;  but  in  the  triangle,  BDC,  BC< 
BD  +  DC.  Taking  EC  from  the  first  member  of  this 
inequality,  and  its  equal,  DC,  from  the  second,  we  have, 
BE  <  BD.  In  the  triangles,  BAE  and  BAD,  BA  is 
common,  and  AE  =  AD  by  construction ;  but  the  third 
side,  BD,  in  the  one,  is  greater  than  the  third  side,  BE, 
in  the  other ;  hence,  the  angle  BAD  is  greater  than  the 
angle  BAE,  (Th.  22,  B.  I);  that  is,  [_BAE  <  [_BAD; 
adding  the  \_EAC  to  the  first  member  of  this  inequality, 
and  its  equal,  th.e.[_DAC,  to  the  other,  we. have 

[_BAE+l_EAC<  [_BAD  +  [_DAC. 
And,  as  the  [_ BAG  is  made  up  of  the  angles  BAE  and 
EAC,  we  have,  as  enunciated, 

]_BAC<  [_BAD  +  [_DAC. 

THEOREM    XIX. 

The  sum  of  the  plane  angles  forming  any  solid  angle,  is 
always  less  than  four  right  angles. 

Let  the  planes  which  form  the  solid  angle  at  A,  be  cut 
by  another  plane,  which  we  may  call  the  plane  of  the 
base,  BCDE.  Take  any  point,  a,  in  this  plane,  and  draw 
aBy  aC,  aD,  aE,  etc.,  thus  making  as  many  triangles  on 


BOOK    VI. 


169 


the  plane  of  the  base  as  there  are  tri- 
angular planes  forming  the  solid  angle 
A.  Now,  since  the  sum  of  the  angles 
of  every  A  is  two  right  angles,  the  sum 
of  all  the  angles  of  the  A's  which 
have  their  vertex  in  A,  is  equal  to  the 
Bum  of  all  angles  of  the  A's  which  have 
their  vertex  in  a.  But,  the  angles  BOA 
+  AOD,  are,  together,  greater  than 
the  angles  BOa  +  aOD,  or  BCD,  by  the  last  proposition. 
That  is,  the  sum  of  all  the  angles  at  the  bases  of  the  A's 
which  have  their  vertex  in  A,  is  greater  than  the  sum  of 
all  the  angles  at  the  bases  of  the  A's  which  have  their 
vertex  in  a.  Therefore,  the  sum  of  all  the  angles  at  a  is 
greater  than  the  sum  of  all  the  angles  at  A ;  but  the  sum 
of  all  the  angles  at  a  is  equal  to  four  right  angles ;  there- 
fore, the  sum  of  all  the  angles  at  A  is  less  than  four  right 
angles. 

THEOREM    XX. 

If  two  solid  angles  are  formed  by  three  plane  angles  respect- 
ively equal  to  each  other,  the  planes  which  contain  the  equal 
angles  will  be  equally  inclined  to  each  other. 

Letthe  [_ASC=the  [_DTF, 
the  |_  ASB  =  the  [_  J>  TE,  and 
the  [_BSO=  the  [_ETF;  then 
will  the  inclination  of  the 
planes,  A  SO,  ASB,  be  equal 
to  that  of  the  planes,  DTF, 
DTE. 

Having  taken  SB  at  pleas- 
are,  draw  BO  perpendicular 

to  the  plane  ASO;  from  the  point  0,  at  which  that  perpen- 
dicular meets  the  plane,  draw  OA  and  00,  perpendicular 
to  SA  and  SO;  draw  AB  and  BO;  next  take  TE  =  SB, 
and  draw  EP  perpendicular  to  the  plane  DTF;  from  the 
15 


170  GEOMETRY. 

point  P,  draw  PB  and  PF,  perpendicular  to  TB  and 
TF;  lastly,  draw  BE  and  FF. 

The  triangle  SAB,  is  right-angled  at  A,  and  the  tri- 
angle TBF,  at  B,  (Th.  5) ;  and  since  the  |__  ASB  =  the 
L  J>  TF,  we  have  [_  SB  A  -  L  2^  J  likewise,  SB=TF; 
therefore,  the  triangle  SAB  is  equal  to  the  triangle  TBF) 
hence,  SA  =  TB,  and  AB  =  Di?.  In  like  manner  it 
may  be  shown  that  SO  =  TF,  and  BO  =  J£F.  That 
granted,  the  quadrilateral  SAOO  is  equal  to  the  quadri- 
lateral TBPF;  for,  place  the  angle  A  SO  upon  its  equal, 
2)2^,  and  because  SA  =  2'i),  and  SO  =  TF,  the  point  A 
will  fall  on  i),  and  the  point  0 on  F;  and,  at  the  same  time, 
A  0,  which  is  perpendicular  to  SA,  will  fall  on  PB,  which 
is  perpendicular  to  TB,  and,  in  like  manner,  00  on  PF; 
wherefore,  the  point  0  will  fall  on  the  point  P,  and  A  0 
will  be  equal  to  BP.  But  the  triangles,  A  OB,  BPF,  are 
right  angled  at  0  and  P;  the  hypotenuse  AB  =  BF,  and 
the  side  AO  =  BP;  hence,  those  triangles  are  equal, 
(Cor,  Th.  39,  B.  I),  and  \_OAB=[_PBF.  The  angle  OAB 
is  the  inclination  of  the  two  planes,  ASB,  ASO;  the  angle 
PBF  is  that  of  the  two  planes,  BTF,  BTF;  conse- 
quently, those  two  inclinations  are  equal  to  each  other. 

Hence  the  theorem. 

Scholium  1.  —  The  angles  which  form  the  solid  angles  at  S  and  T, 
may  be  of  such  relative  magnitudes,  that  the  perpendiculars,  BO  and 
EP,  may  not  fall  within  the  bases,  ASC  and  DTF;  but  they  will 
always  either  fall  on  the  bases,  or  on  the  planes  of  the  bases  produced, 
and  0  will  have  the  same  relative  situation  to  A,  S,  and  C,  as  P  has 
to  D,  T,  and  F.  In  case  that  0  and  P  fall  on  the  planes  of  the  bases 
produced,  the  angles  BCO  and  EFP,  would  be  obtuse  angles ;  but  the 
demonstration  of  the  problem  would  not  be  varied  in  the  least. 

Scholium  2.  —  If  the  plane  angles  bounding  one  of  the  triedral 
angles  be  equal  to  those  of  the  other,  each  to  each,  and  also  be  simi 
larly  arranged  about  the  triedral  angles,  these  solid  angles  will  be  ab- 
solutely equal.  For  it  was  shewn,  in  the  course  of  the  above  demon- 
stration, that  the  quadrilaterals,  SAOC and  TDPF,  were  equal;  and 
on  being  applied,  the  point  0  falls  on  the  point  P;  and  since  the  trian- 
gles A  OB  and  DPE  are  equal,  the  perpendiculars  OB  and  PE  ar« 


BOOK  VI.  171 

ilsc  equal.  Now,  because  the  plane  angles  are  like  arranged  about 
the  triedral  angles,  these  perpendiculars  lie  in  the  same  direction ; 
hence  the  point  B  will  fall  on  the  point  \E,  and  the  solid  angles 
will  exactly  coincide. 

Scholium  3. — When  the  planes  of  the  equal  angles  are  not  like  dis- 
posed about  the  triedral  angles,  it  would  not  be  possible  to  make  these 
triedral  angles  coincide  ;  and  still  it  would  be  true  that  the  planes  of 
the  equal  angles  are  equally  inclined  to  each  other.  Hence,  these 
triedral  angles  have  the  plane  and  diedral  angles  of  the  one,  equal  to 
the  plane  and  diedral  angles  of  the  other,  each  to  each,  without  having 
of  themselves  that  absolute  equality  which  admits  of  superposition. 
Magnitudes  which  are  thus  equal  in  all  their  component  parts,  but 
will  not  coincide,  when  applied  the  one  to  the  other,  are  said  to  be 
symmetrically  equal.  Thus,  two  triedral  angles,  bounded  by  plane 
fcngbs  equal  each  to  each,  but  not  like  placed,  are  symmetrical  triedral 


172  GEOMETRY. 


BOOK  VII, 


SOLID    GEOMETRY. 
DEFINITIONS. 

1.  A  Polyedron  is  a  solid,  or  volume,  bounded  on  all 
sides  by  planes.  The  bounding  planes  are  called  the 
faces  of  the  polyedron,  and  then  intersections  are  its 
edges, 

2.  A  Prism  is  a  polyedron,  having  two  of  its  faces, 
called  bases,  equal  polygons,  whose  planes  and  homolo- 
gous sides  are  parallel.  The  other,  or  lateral  faces,  are 
parallelograms,  and  constitute  the  convex  surface  of  the 
prism. 

The  bases  of  a  prism  are  distinguished  by  the  terms, 
upper  and  lower;  and  the  altitude  of  the  prism  is  the  per 
pendicular  distance  between  its  bases. 

Prisms  are  denominated  triangular,  quadrangular,  pent 
angular,  etc.,  according  as  their  bases  are  triangles,  quad- 
rilaterals, pentagons,  etc. 

3.  A  Right  Prism  is  one  in  which  the  planes  of  the 
lateral  faces  are  perpendicular  to  the  planes  of  the  bases. 

4.  A  Parallelopipedon  is  a  prism 
whose  bases  are  parallelograms. 

5.  A  Rectangular  Parallelopipedon 
is  a  right  parallelopipedon,  with 
rectangular  bases. 


BOOK   VII. 


173 


6.  A  Cube  or  Hexaedron  is  a  rectangu- 
lar parallelopipedon,  whose  faces  are  all 
equal  squares. 

7.  A  Diagonal  of  a  Polyedron  is  a  straight 
line  joining  the  vertices  of  two  solid 
angles  not  adjacent. 

8.  Similar  Polyedrons  are  those  which 

are  bounded  by  the  same  number  of  similar  polygons 
like  placed,  and  whose  homologous  solid  angles  are 
equal. 

Similar  parts,  whether  faces,  edges,  diagonals,  or 
angles,  similarly  placed  in  similar  polyedrons,  are  termed 
homologous. 

9.  A  Pyramid  is  a  polyedron,  having 
for  one  of  its  faces,  called  the  base,  any 
polygon  whatever,  and  for  its  other  faces 
triangles  having  a  common  vertex,  the 
sides  opposite  which,  in  the  several  trian- 
gles, being  the  sides  of  the  base  of  the 
pyramid. 

10.  The  Vertex  of  a  pyramid  is  the 
common  vertex  of  the  triangular  faces. 

U.  The  Altitude  of  a  pyramid  is  the  perpendicular 
distance  from  its  vertex  to  the  plane  of  its  base. 

12.  A  Right  Pyramid  is  one  whose  base  is  a  regular 
polygon,  and  whose  vertex  is  in  the  perpendicular  to  the 
base  at  its  center.  This  perpendicular  is  called  the  axis 
of  the  pyramid. 

13.  The  Slant  Height  of  a  right  pyramid  is  the  perpen- 
dicular distance  from  the  vertex  to  one  of  the  sides  of 
the  base. 

14.  The  Frustum  of  a  Pyramid  is  a  portion  of  the  pyr- 
amid included  between  its  base  and  a  section  made  by  a 
plane  parallel  to  the  base. 

Pyramids,  lik*  prisms,  are  named  from  the  forms  of 
their  bases. 
15* 


174 


GEOMETRY. 


15.  A  Cylinder  is  a  body,  having  for 
its  ends,  or  bases,  two  equal  circles, 
the  planes  of  which  are  perpendicular 
to  the  line  joining  their  centers ;  the 
remainder  of  its  surface  may  be  con- 
ceived as  formed  by  the  motion  of  a 
line,  which  constantly  touches  the  cir- 
cumferences of  the  bases,  while  it 
remains  parallel  to  the  line  which 
joins  their  centers. 

We  may  otherwise  define  the  cylinder  as  a  body  gen- 
erated by  the  revolution  of  a  rectangle  about  one  of  its 
sides  as  an  immovable  axis. 

The  sides  of  the  rectangle  perpendicular  to  the  axis 
generate  the  bases  of  the  cylinder;  and  the  side  opposite 
the  axis  generates  its  convex  surface.  The  line  joining 
the  centers  of  the  bases  of  the  cylinder  is  its  axis,  and  is 
also  its  altitude. 

If,  within  the  base  of  a  cylinder,  any  polygon  be  in- 
scribed, and  on  it,  as  a  base,  a  right  prism  be  con- 
structed, having  for  its  altitude  that  of  the  cylinder,  such 
prism  is  said  to  be  inscribed  in  the  cylinder,  and  the  cylin- 
der is  said  to  circumscribe  the  prism. 

Thus,  in  the  last  figure,  ABODEc  is  an  inscribed 
prism,  and  it  is  plain  that  all  its  lateral  edges  are  con- 
tained in  the  convex  surface  of  the  cylinder.       d 

If,  about  the  base  of  a  cylinder,  any 
polygon  be  circumscribed,  and  on  it, 
as  a  base,  a  right  prism  be  con- 
structed, having  for  its  altitude  that 
of  the  cylinder,  such  prism  is  said  to 
be  circumscribed  about  the  cylinder,  and 
the  cylinder  is  said  to  be  inscribed  in 
the  prism. 

Thus,  ABCLEFc  is  a  circum- 
scribed prism;   and  it  is  plain  that 


BOOK    Vll. 


175 


the  lino,  mn,  which  joins  the  points  of 
tangency  of  the  sides,  EF  and  ef,  with 
the  circumferences  of  the  bases  of  the 
cylinder,  is  common  to  the  convex  sur- 
faces of  the  cylinder  and  prism. 

16.  A  Cone  is  a  body  bounded  by  a 
circle  and  the  surface  generated  by  the 
motion  of  a  straight  line,  which  con- 
stantly passes  through  a  point  in  the 
perpendicular  to  the  plane  of  the  circle 
at  its  center,  and  the  different  points  in 
its  circumference. 

The  cone  may  be  otherwise' defined  as  a  body  gene 
rated  by  the  revolution  of  a  right-angled  triangle  about 
one  of  its  sides  as  an  immovable  axis.     The  other  side 
of  the  triangle  will  generate  the  base  of  the  cone,  while 
the  hypotenuse  generates  the  convex  surface. 

The  side  about  which  the  generating  triangle  revolves 
is  the  axis  of  the  cone,  and  is  at  the  same  time  its  altitude. 

If,  within  the  base  of  the  cone,  any 
polygon  be  inscribed,  and  on  it,  as  a 
base,  a  pyramid  be  constructed,  having 
for  its  vertex  that  of  the  cone,  such 
pyramid  is  said  to  be  inscribed  in  the 
cone,  and  the  cone  is  said  to  circumscribe 
the  pyramid. 

Thus,  in  the  accompanying  figure, 
V — ABODE,  is  an  inscribed  pyramid, 
and  it  is  plain  that  all  its  lateral  edges 
are  contained  in  the  convex  surface  of 
the  cone. 

If,  about  the  base  of  a  cone,  any  poly- 
gon be  circumscribed,  and  on  it,  as  a 
base,  a  pyramid  be  constructed,  having 
for  its  vertex  that  of  the  cone,  such  pyramid  is  said  to  be 
circumscribed  about  the  cone,  and  the  cone  is  said  to  be 
inscribed  in  the  pyramid. 


176 


GEOMETRY. 


17.  The  Frnstnm  of  a  Cone  is  the  portion  of  the  cone  that 
is  included  between  its  base  and  a  section  made  by  a  plane 
parallel  to  the  base. 

18.  Similar  Cylinders,  and  also  Similar  Cones,  are  such  as 
have  their  axes  proportional  to  the  radii  of  their  bases. 

19.  A  Sphere  is  a  body  bounded  by  one  uniformly-curved 
surface,  all  the  points  of  which  are  at  the  same  distance 
from  a  certain  point  within,  called  the  center. 

We  may  otherwise  define  the  sphere  as  a  body  gene- 
rated by  the  revolution  of  a  semicircle  about  its  diameter 
as  an  immovable  axis. 

20.  A  Spherical  Sector  is  that 
portion  of  a  sphere  which  is  in- 
cluded between  the  surfaces  of 
two  cones  having  a  common 
axis,  and  their  vertices  at  the 
center  of  the  sphere.  Or,  it  is 
that  portion  of  the  sphere  which 
is  generated  by  a  sector  of  the 
generating  semicircle. 

21.  The  Radius  of  a  Sphere  is 
a  straight  line  drawn  from  the 

center  to  any  point  in  the  surface ;  and  the  diameter  is 
a  straight  line  drawn  through  the  center,  and  limited  on 
both  sides  by  the  surface. 

All  the  diameters  of  a  sphere  are  equal,  each  being 
twice  the  radius. 

22.  A  Tangent  Plane  to  a  sphere  is  one  which  has  a 
single  point  in  the  surface  of  the  sphere,  all  the  others 
being  without  it. 

23.  A  Secant  Plane  to  a  sphere  is  one  which  has  more 
than  one  point  in  the  surface  of  the  sphere,  and  lies 
partly  within  and  partly  without  it. 

Assuming,  what  will  presently  be  proved,  that  the  in- 
tersection of  a  sphere  by  a  plane  is  a  circle, 

24.  A  Small  Circle  of  a  sphere  is  one  whose  plane  does 
not  pass  through  its  center;  and 


BOOK   VII 


177 


25.  A  Great  Circle  of  a  sphere  is  one  whose  plane  passes 
through  the  center  of  the  sphere. 

26.  A  Zone  of  a  sphere  is  the  portion  of  its  surface  in- 
cluded between  the  circumferences  of  any  two  of  its  paral- 
lel circles,  called  the  bases  of  the  zone.  "When  the  plane 
of  one  of  these  circles  becomes  tangent  to  the  sphere,  the 
zone  has  a  single  base. 

27.  A  Spherical  Segment  is  a  portion  of  the  volume  of  a 
sphere  included  between  any  two  of  its  parallel  circles, 
called  the  bases  of  the  segment. 

The  altitude  of  a  zone,  or  of  a  segment,  of  a  sphere, 
is  the  perpendicular  distance  between  the  planes  of  its 
bases.  \ 

28.  The  area  of  a  surface  is  measured  by  the  product 
of  its  length  and  breadth,  and  these  dimensions  are  always 
conceived  to  be  exactly  at  right  angles  to  each  other. 

29.  In  a  similar  manner,  solids  are  measured  by  the 
product  of  their  length,  breadth,  and  height,  when  all  their 
dimensions  are  at  right  angles  to  each  other. 

The  product  of  the  length  and  breadth  ©f  a  solid,  is 
the  measure  of  the  surface  of  its  base. 

Let  P,  in  the  annexed  fig- 
ure, represent  the  measuring 
unit,  and  AF  the  rectangular 
solid  to  be  measured. 

A  side  of  P  is  one  unit  in 
length,  one  in  breadth,  and 
one  in  height;  one  inch,  one 
foot,  one  yard,  or  any  other  unit  that  may  be  taken. 


Then, 


lxlxl 


1,  the  unit  cube. 


Now,  if  the  base  of  the  solid,  AC,  is,  as  here  repr<  • 
sented,  5  units  in  length  and  2  in  breadth,  it  is  obvious 
that  (5x2  =  10),  10  units,  each  equal  to  P,  can  be  placed 
on  the  base  of  AC,  and  no  more;  and  as  each  of  these 
onits  will  occupy  a  unit  of  altitude,  therefore,  2  units  of 

M 


178 


GEOMETRY. 


altitude  will  contain  20  solid  units,  3  units  of  altitude, 
30  solid  units,  and  so  on ;  or,  in  general  terms,  the  num- 
ber of  square  units  in  the  base  multiplied  by  the  linear  units 
in  perpendicular  altitude,  will  give  the  solid  units  in  any  rect- 
angular solid. 


THEOREM    I. 

If  the  three  plane  faces  bounding  a  solid  angle  of  one  prism 
be  equal  to  the  three  plane  faces  bounding  a  solid  angle  of 
another,  each  to  each,  and  similarly  disposed,  the  prisms  will 
be  equal. 

Suppose  A  and  a  to  be  the  vertices  of  two  solid  angles, 
bounded  by  equal  and  similarly  placed  faces;  then  will 
the  prisms,  ABODE— -i^and  abcde — n,  be  equal. 

For,  if  we  place  the  base, 
abcde,  upon  its  equal,  the  base 
ABODE,  they  will  coincide; 
and  since  the  solid  angles, 
whose  vertices  are  A  and  a,  are 
equal,  the  lines  ab,  ae,  and  ap, 
respectively  coincide  with  AB, 
AE,  and  AP ;  but  the  faces,  al  and  ao,  of  the  one  prism, 
are  equal,  each  to  each,  to  the  faces,  AL  and  AO,  of  the 
other;  therefore  pi  and  po  coincide  with  PL  and  PO, 
and  the  upper  bases  of  the  prisms  also  coincide :  hence, 
not  only  the  bases,  but  all  the  lateral  faces  of  the  two 
prisms  coincide,  and  the  prisms  are  equal. 

Oor.  If  the  two  prisms  are  right,  and  have  equal  bases 
and  altitudes,  they  are  equal.  For,  in  this  case,  the  rect- 
angular faces,  al  and  ao,  of  the  one,  are  respectively 
equal  to  the  rectangular  faces,  AL  and  AO,  of  the  other; 
and  hence  the  three  faces  bounding  a  triedral  angle  in 
the  one,  are  equal  and  like  placed,  to  the  faces  bounding 
a  triedral  angle  in  the  othei 


BOOK    VII.  179 

THEOREM    II. 

The  opposite  faces  of  any  parattelopipedon  axe  equal,  and 
their  planes  are  parallel. 

Let  ABOD — E  be  any  parallelopipedon ;  then  will  its 
opposite  faces  be  equal,  and  their  planes  will  be  parallel. 

The  bases  ABOD  and  FEGH  are 
equal,  and  their  planes  are  parallel, 
by  definitions  2  and  4  of  this  Book ; 
it  remains  tor  us,  therefore,  only  to 
show  that  any  two  of  the  opposite 
lateral  faces  are  equal  and  parallel. 

Since  all  the  faces  of  the  parallelopipedon  are  parallel- 
ograms, AB  is  equal  and  parallel  to  DC,  and  AH  is  also 
equal  and  parallel  to  DF;  hence  the  angles  HAB  and 
FDO  are  equal,  and  their  planes  are  parallel,  (Th.  17,  B. 
VI),  and  the  two  parallelograms,  HABGr  and  FDCE, 
having  two  adjacent  sides  and  the  included  angle  of  the 
one  equal  to  the  two  adjacent  sides  and  included  angle 
of  the  other,  are  equal. 

Oor.  1  Hence,  of  the  six  faces  of  the  parallelopipedon, 
any  two  lying  opposite  may  be  taken  as  the  bases. 

Cor.  2.  The  four  diagonals  of  a  parallelopipedon  mutu- 
ally bisect  each  other.  For,  if  we  draw  A  0  and  HE,  we 
shall  form  the  parallelogram  A  OEH,  of  which  the  diago- 
nals are  AE  and  HO,  and  these  diagonals  are  at  the  same 
time  diagonals  of  the  parallelopipedon;  but  the  diagonals 
of  a  parallelogram  mutually  bisect  each  other.  Now,  if 
the  diagonal  FB  be  drawn,  it  and  HO  will  bisect  each 
other,  since  they  are  diagonals  'of  the  parallelogram 
FHBO.  In  like  manner  we  can  show  that  if  DGr  be 
drawn,  it  will  be  bisected  by  AE.  Hence,  the  four  diag- 
onals have  a  common  point  within  the  parallelopipedon. 

Scholium.  —  It  is  seen  at  once  that  the  six  faces  of  a  parallelopipe- 
don intersect  each  other  in  twelve  edges,  four  of  which  are  equal  to 
HA,  four  to  AB,  and  four  to  AD.  Now,  we  may  conceive  the  parallel- 
opipedon to  be  bounded  by  the  planes  determined  by  the  three  lines 


180 


GEOMETR 


AH,  AB,  and  AD,  and  the  three  planes  passed  through  the  extremi* 
ties,  H,  B,  and  JD,  of  these  lines,  parallel  to  the  first  three  planes. 


THEOREM    III. 

The  convex  surface  of  a  right  prism  is  measured  by  the 
perimeter  of  its  base  multiplied  by  its  altitude. 

Let  ABODE — iVbe  a  right  prism,  of 
which  AP  is  the  altitude ;  then  will  its 
convex  surface  he  measured  by 

{AB  +  BO  +  OB  +  DE  +  EA)xAP. 
For,  its  convex  surface  is  made  up  of  the 
rectangles  AL,  BM,  ON,  etc.,  and  each 
rectangle  is  measured  by  the  product  of 
its  base  by  its  altitude ;  but  the  altitude 
of  each  rectangle  is  equal  to  AP,  the  alti- 
tude of  the  prism ;  hence  the  convex  sur- 
face of  the  prism  is  measured  by  the  pro- 
duct of  the  sum  of  the  bases  of  the  rectangles,  or  the 
perimeter  of  the  base  of  the  prism,  by  the  common  alti- 
tude, AP. 

Oor.  Kight  prisms  will  have  equivalent  convex  surfaces, 
when  the  products  of  the  perimeters  of  their  bases  by 
their  altitudes  are  respectively  equal ;  and,  generally,  their 
convex  surfaces  will  be  to  each  other  as  the  products  of 
the  perimeters  of  their  bases  by  their  altitudes.  Hence, 
if  the  altitudes  are  equal,  their  convex  surfaces  will  be  as 
the  perimeters  of  their  bases ;  and  if  the  perimeters  of 
their  bases  are  equal,  the^r  convex  surfaces  will  be  as 
their  altitudes. 


THEOREM   IV. 


The  two  sections  of  a  prism  made  by  parallel  planes  between 
its  bases  are  equal  polygons. 

Let  the  prism  ABODE — N  be  cut  between  its  bases 
by  two  parallel  planes,  making  the  sections  QBS,  etc.. 


BOOK   VII, 


181 


and  TVX,  etc. ;  then  will  these  sections 
be  equal  polygons. 

For,  since  the  secant  planes  are  paral- 
lel, their  intersections,  QR  and  TV,  by 
the  plane  of  the  face  JEAPO  are  parallel, 
(Th.  9,  B.  VI) ;  and  being  included  be- 
tween the  parallel  lines,  AP  and  EO,  they 
are  also  equal.  In  the  same  manner  we 
may  prove  that  BS  is  equal  and  parallel 
to  VX,  and  so  on  for  the  intersections  of 
the  secant  planes  by  the  other  faces  of 
the  prism.  Hence,  these  polygonal  sections  have  the 
sides  of  the  one  equal  to  the  sides  of  the  other,  each  to 
each.  The  angles  QBS  and  TVX  are  equal,  because 
their  sides  are  parallel  and  lie  in  the  same  direction ;  and 
in  like  manner  we  prove  [__  RSY  =  [_  VXZ,  and  so  on 
for  the  other  corresponding  angles  of  the  polygons. 
Therefore,  these  polygons  are  both  mutually  equilateral 
and  mutually  equiangular,  and  consequently  are  equal. 

Cor.  A  section  of  a  prism  made  by  a  plane  parallel  to 
the  base  of  the  prism,  is  a  polygon  equal  to  the  base. 


THEOKEM   V. 

Two  parallelopipedons,  the  one  rectangular  and  the  other 
oblique,  will  be  equal  in  volume  when,  having  the  same  base 
and  altitude,  two  opposite  lateral  faces  of  the  one  are  m  the 
'planes  of  the  corresponding  lateral  faces  of  the  other. 

Designating  the  parallelo- 
pipedons by  their  opposite 
diagonal  letters,  let  AG  be 
the  rectangular,  and  AL  the 
obnque,  parallelopipedon,  hav- 
ing the  same  base,  A  C,  and 
the  same  altitude,  namely, 
the  perpendicular  distance  be- 
16 


182  GEOMETRY. 

tween  the  parallel  pranes,  AC  and  EL.  Also  let  the  fa^e, 
AK,  be  in  the  plane  of  the  face,  AF,  and  the  face,  BL,  m 
the  plane  of  the  face,  BG.  We  are  now  to  prove  that  the 
oblique  parallelopipedon  is  equivalent  to  tne  rectangular 
parallelopipedon. 

As  the  faces,  AF  and  AK,  are  in  the  same  plane,  and 
the  parallelopipedons  have  the  same  altitude,  EFK  is  a 
straight  line,  and  EF— IK,  because  each  is  equal  to  AB. 
If  from  the  whole  line,  FK,  we  take  FF,  and  then  from 
the  same  line  we  take  IK=  FF,  we  shall  have  the  re- 
mainders, FI  and  FK,  equal ;  and  since  AF  and  BF  are 
parallel,  \_AFI  =  [_BFK ;  hence  the  A's,  AFI  and 
BFK,  are  equal.  Since  HE  and  MI  are  both  parallel  to 
DA,  they  are  parallel  to  each  other,  and  FIMH  is  a  par- 
allelogram ;  for  like  reasons,  FKLG  is  a  parallelogram, 
and  these  parallelograms  are  equal,  because  two  adjacent 
sides  and  the  included  angle  of  the  one  are  equal  to  two 
adjacent  sides  and  the  included  angle  of  the  other.  The 
parallelograms,  BE  and  CF,  being  the  opposite  faces  of 
the  parallelopipedon,  AG-,  are  equal.  Hence,  the  three 
plane  faces  bounding  the  triedral  angle,  E,  of  the  triaii* 
gular  prism,  EAI — E,  are  equal,  each  to  each,  and  like 
placed,  to  the  three  plane  faces  bounding  the  triedral  angle 
F,  of  the  triangular  prism,  FBK —  G,  and  these  prisms 
are  therefore  equal,  (Th.  1).  Now,  if  from  the  whole 
solid,  EABK — H,  we  take  the  prism,  EAI — H,  there 
will  remain  the  parallelopipedon,  AL;  and,  if  from  the 
srme  solid,  we  take  the  prism,  FBK—G,  there  will  remain 
the  rectangular  parallelopipedon,  AG.  Therefore,  the 
oblique  and  the  rectangular  parallelopipedons  are  equiva- 
lent. 

Cor.  The  volume  of  the  rectangular  parallelopipedon, 
AG,  is  measured  by  the  base,  ABCB,  multiplied  by  the 
altitude,  AF,  (Def.  29) ;  consequently,  the  oblique  paral- 
lelopipedon is  measured  by  the  product  of  the  same  base 
by  the  same  altitude. 


BOOK   VII. 


183 


Scholium.-  -If  neither  of  the  parallelopipedons  is  rectangular,  but 
they  still  have  the  same  base  and  the  same  altitude,  and  two  opposite 
lateral  faces  of  the  one  are  in  the  planes  of  the  corresponding  lateral 
faces  of  the  other,  by  precisely  the  same  reasoning  we  could  prove  the 
parallelopipedons  equivalent.  Hence,  in  general,  any  two  parallelo- 
pipedons will  be  equal  in  volume  when,  having  the  same  base  and  aUtiPude, 
two  opposite  lateral  faces  of  the  one  are  in  the  planes  of  the  correspond- 
ing lateral  faces  of  the  other, 

THEOREM    VI. 

Two  parallelopipedons  having  equal  bases  and  equal  alti- 
tudes^  are  equivalent. 

Let  A  G-  and  AL  be  two  paral- 
lelopipedons, having  a  common 
lower  base,  and  their  upper  bases 
in  the  same  plane,  HF.  Then 
will  these  parallelopipedons  be 
equivalent. 

Since  their  upper  bases  are  in 
the  same  plane,  and  the  lines  IM  and  KL  are  par- 
allel, and  also  EF  and  HG,  these  lines  will  intersect, 
when  produced,  and  form  the  parallelogram  NOPQ, 
which  will  be  equal  to  the  common  lower  base  of 
the  two  parallelopipedons.  Now,  if  a  third  parallelo- 
pipedon  be  constructed,  having  BD  for  its  lower  base, 
and  OQ  for  its  upper  base,  it  will  be  equivalent  to  the  par- 
allelopipedon  AG,  and  also  to  the  parallelopipedon  AL, 
(Th.  5,  Scholium) ;  hence,  the  two  given  parallelopipe- 
dons, being  each  equivalent  to  the  third  parallelopipe- 
don, are  equivalent  to  each  other. 

Hence,  two  parallelopipedons  having  equal  bases,  etc. 


THEOREM    VII. 

The  volume  of  any  parallelopipedon  is  measured  by  the 
product  of  its  base  and  altitude,  or  the  product  of  its  three 
dimensions. 


184  GEOMETRY. 

Let  ABCD—Gr  be  any  parallelopipedon ;  tl  en  will  its 
volume  be  expressed  by  the  product       ^_h 


of  tbe  area  of  its  base  and  altitude.  hj 

If  the  parallelopipedon  is  oblique,     l/ 
we  may  construct  on  its  base  a  right 
parallelopipedon,  by  erecting  perpen- 
diculars at  the  points  A,  B,  0,  and  D, 
and  making  them  each  equal  to  the 
altitude  of  the  given  parallelopipedon ; 
and  the  right  parallelopipedon,  thus 
constructed,  will  be  equivalent  to  the  given  parallelopip- 
edon, (Th.  6).     Eow,  if  the  base,  ABOD,  is  a  rectangle, 
the  new  parallelopipedon  will  be  rectangular,  and  meas- 
ured by  the  product  of  its  base  and  altitude,  (Def.  29). 
But  if  the  base  is  not  rectangular,  let  fall  the  perpen- 
diculars, Be  and  Ad,  on  CD  and  CD  produced,  and  take 
the  rectangle  ABcd  for  the  base  of  a  rectangular  paral- 
lelopipedon, having  for  its  altitude  that  of  the  given 
parallelopipedon.     "We  may  now  regard  the  rectangular 
face,  ABFE,  as  the  common  base  of  the  two  parallelo- 
pipedons,  Ag  and  AG-',  and,  as  they  have  a  common 
base,  and  equal  altitude,  they  are  equivalent.     Thus  we 
have  reduced  the  oblique  parallelopipedon,  first  to  an 
equivalent  right  parallelopipedon  on  the  same  base,  and 
then  the  right  to  an  equivalent  rectangular  parallelopip- 
edon on  an  equivalent  base,  all  having  the  same  alti- 
tude.    But  the    rectangular  parallelopipedon,   Ag,   is 
measured  by  product  of  its  base,  ABcd,  and  its  altitude; 
hence,  the  given  and  equivalent  oblique  parallelopipedon 
is  measured  by  the  product  of  its  equivalent  base  and 
equal  altitude. 

Hence,  the  volume  of  any  parallelopipedon,  etc. 

Cor.  Since  a  parallelopipedon  is  measured  by  the  pro- 
duct of  its  base  by  its  altitude,  it  follows  that  parallelo 
pipedons  of  equivalent  bases,  and  equal  altitude?,  are  equiva 
lent,  or  equal  in  volume. 


BOOK    VII  185 

THEOREM   VIII. 

Parallelopipedons  on  the  same,  or  equivalent  hoses,  are  to 
each  other  as  their  altitudes ;  and  parallelopipedons  having 
equal  altitudes,  are  to  each  other  as  their  bases. 

Let  P  and  p  represent  two  parallelopipedons,  whose 
bases  are  denoted  by  B  and  b,  and  altitudes  by  A  and  a, 
respectively. 

Now,    P  =  B  x  A,  and  p  =  b  x  a,  (Th.  7). 

But  magnitudes  are  proportional  to  their  numerical 
measures ;  that  is, 

P  :  p  ::  B  x  A  :  b  X  a. 

If  the  bases  of  the  parallelopipedons  are  equivalent, 
we  have  B  —  b;  and  if  the  altitudes  are  equal,  we  have 
A  =  a.  Introducing  these  suppositions,  in  succession, 
in  the  above  proportion,  we  get 

P  :  p  ::  A  :  a, 
and  P  :  p  : :  B  :  b. 

Hence  the  theorem ;  Parallelopipedons  on  the  same,  etc, 

THEOREM    IX. 

Similar  parallelopipedons  are  to  each  other  as  the  cubes  of 
their  like  dimensions. 

Let  P  and  p  represent  any  two  similar  parallelopipe- 
dons, the  altitude  of  the  first  being  denoted  by  h,  and 
the  length  and  breadth  of  its  base  by  I  and  n,  respect- 
ively; and  let  h',  V,  and  nr,  in  order,  denote  the  correeh 
ponding  dimensions  of  the  second. 
Then  we  are  to  prove  that 

P  :  p  ::  n8  :  n"  ::  I*  :  V*  ::  hn  :  h'\ 
We  have 

P  =  Inh,  and  p  -  Vn'K*  (Th.  7) ; 
and  by  dividing  the  first  of  these   equations  by  the 
second,  member  by  member,  we  get 
16* 


186  GEOMETRY. 

P  _   Inh 
J~~  I'n'h'' 
which,  reduced  to  a  proportion,  gives 
P  :  p  ::  Inh  :  Vn'Jt. 
But,  by  reason  of  the  similarity  of  the  patallelopipo 
dons,  we  have  the  proportions 

I  if  : ;  n  :  n' 
h  :  h1  : :  n  :  n' ; 
we  have  also  the  identical  proportion, 

n  :  n'  : :  n  :  n' . 
By  the  multiplication  of  these  proportions,  term  by 
term,  we  get,  (Th.  11,  B.  II), 

Inh  :  l'nfhf  : :  n*  :  nn. 
That  is,  P  :  p  : :  n*  :  ri\ 

By  treating  in  the  same  manner  the  three  proportions, 
I  :  V  : :  h  :  h' 
n  :  nr  :  :  h  :  hr 
h  :  h'  : :  h  :  hf, 
we  should  obtain  the  proportion 

P  :  p  ::  h*  :  h"; 
and,  by  a  like  process,  the  three  proportions, 
h  :  h'  :  :  I  :  V 
n  :  n'  : :  I  :  V 
I  :  V  ::  I  :  I', 
will  give  us  the  proportion 

P  :  p  ::  I*  :  V\ 
Hence  the  theorem ;  similar  parallelopipedons  are  to  each 
other,  etc. 

THEOREM   X. 

The  two  triangular  prisms  into  which  any  parallelopipedon 
is  divided,  by  a  plane  passing  through  its  opposite  diagonal 
edges,  are  equivalent. 

Let  ABCD — F  be  a  parallelopipedon,  and  through 
the  diagonal  edges,  BF  and  DH,  pass  the  plane  BH.  divi- 
ding the  parallelopipedon  into  the  two  triangular  prisms. 


BOOK    VII. 


187 


ABB-E  and  BCB—G ;  then  we  are  to  prove  that  these 
prisms  an  equivalent.  Letus  divide 
the  diagonal,  BB,  in  which  the  se- 
cant plane  intersects  the  base  of  the 
parallelopipedon,  into  three  equal 
parts,  a  and  c  being  the  points  of 
division.  In  the  base,  AB  CD,  con- 
st ruct  the  complementary  paral- 
lelograms, aC  and  a  A,  and  in  the 
parallelogram,  badD,  construct  the 
complementary  parallelograms, 
cd  and  cb,  and  conceive  these,  to- 
gether with  the  parallelograms, 
Ba,  ac,  cB,  to  be  the  bases  of 
smaller  parallelopipedons,  having 
their  lateral  faces  parallel  to  the 

lateral  faces  of,  and  their  altitude  equal  to  the  altitude  of, 
the  given  parallelopipedon,  A  Gr. 

Now  it  is  evident  that  the  triangular  prism,  BCD — Gr, 
is  composed  of  the  parallelopipedons  on  the  bases,  a  0 
and  cd,  and  the  triangular  prisms,  on  the  side  of  the 
secant  plane  with  this  prism,  into  which  this  plane  divides 
the  parallelopipedons  on  the  bases,  Ba,  ac,  and  cB.  The 
triangular  prism,  ABB — E,  is  also  composed  of  the  par- 
allelopipedons on  the  bases,  Aa  and  be,  together  with  the 
triangular  prisms  on  the  side  of  the  secant  plane  with 
this  prism,  into  which  this  plane  divides  the  parallelopip- 
edons on  the  bases,  Ba,  ac,  and  cD. 

But  the  parallelograms,  a  0  and  a  A,  being  complement- 
ary, are  equivalent,  (Th.  31,  B.  I) ;  and  for  the  same 
reason  the  parallelograms,  cd  and  cb,  are  equivalent ;  and 
since  parallelopipedons  on  equivalent  bases  and  of  equal 
altitudes,  are  equivalent,  (Cor.,  Th.  7),  we  have  the  sum 
of  parallelopipedons  on  bases  a  0  and  cd,  equivalent  to 
the  sum  of  parallelopipedons  on  the  bases,  aA  and  cb. 
Hence,  the  triangular  prisms,  ABB — E  and  BOB — Gr, 


188  GEOMETRY. 

differ  in  volume  only  by  the  difference  which  may  exist 
between  the  sums  of  the  triangular  prisms  on  the  two 
Bides  of  the  secant  plane  into  which  this  plane  divides 
the  parallelopipedons  on  the  bases,  Ba,  ac,  and  cd. 

Now,  if  the  number  of  equal  parts  into  which  the  diag- 
onal is  divided,  be  indefinitely  multiplied,  it  still  holds 
true  that  the  triangular  prisms,  ABB — B  and  BOB — 6r, 
differ  in  volume  only  by  the  difference  between  the  sums 
of  the  triangular  prisms  on  the  two  sides  of  the  secaLt 
plane  into  which  this  plane  divides  the  parallelopipedons 
constructed  on  the  bases  whose  diagonals  are  the  equal 
portions  of  the  diagonal,  BB.  But  in  this  case  the  sum 
of  these  parallelopipedons  themselves  becomes  an  indefi- 
nitely small  part  of  the  whole  parallelopipedon,  A  Cr,  and 
the  difference  between  the  parts  of  an  indefinitely  small 
quantity  must  itself  be  indefinitely  small,  or  less  than 
any  assignable  quantity.  Therefore,  the  triangular 
prisms,  ABB — B  and  BOB — Gr,  differ  in  volume  by  less 
than  any  assignable  volume,  and  are  consequently  equiv- 
alent. 

Hence  the  theorem ;  the  two  triangular  prisms  into  which, 
etc. 

Cor.  1.  Any  triangular  prism,  as  ABB — B,  is  one  half 
the  parallelopipedon  having  the  same  triedral  angle,  A, 
and  the  same  edges,  AB,  AB,  and  AB. 

Cor.  2.  Since  the  volume  of  a  parallelopipedon  is  meas- 
ured by  the  product  of  its  base  and  altitude,  and  the  tri- 
angular prisms  into  which  it  is  divided  by  the  diagonal 
plane,  have  bases  equivalent  to  one  half  the  base  of  the 
parallelopipedon,  and  the  same  altitude,  it  follows  that, 
the  volume  of  a  triangular  prism  is  measured  by  the  product 
of  its  base  and  altitude. 

The  above  demonstration  is  less  direct,  but  is  thought 
to  be  more  simple,  than  that  generally  found  in  authors, 
and  which  is  here  given  aa  a 


BOOK    VII 


189 


Second  Demonstration 

Let  ABCD — F  be  a  parallelo- 
pipedon,  divided  by  the  diagonal 
plane,  BH,  passing  through  the 
edges,  BF  and  BR;  then  we  are 
to  prove  that  the  triangular 
prisms,  ABB—E  and  BQD—G, 
thus  formed,  are  equivalent. 

Through  the  points  B  and  F, 
pass  planes  perpendicular  to  the 
edge,  BF,  and  produce  the  late- 
ral faces  of  the  parallelopipedon 
to  intersect  the  plane  through  B ; 
then  the  sections  Bcda  and  Fghe 
are  equal  parallelograms.  For,  since  the  cutting  planea 
are  both  perpendicular  to  BF,  they  are  parallel,  (Th.  10, 
B.  VI) ;  and  because  the  opposite  faces  of  a  parallelo- 
pipedon are  in  parallel  planes,  (Th.  2),  and  the  intersec- 
tions of  two  parallel  planes  by  a  third  plane  are  parallel, 
(Th.  9,  B.  VI),  the  sections,  Bcda  and  Fghe,  are  equal 
parallelograms,  and  may  be  taken  as  the  bases  of  the 
right  parallelopipedon,  Bcda — h.  But  the  diagonal  plane 
divides  the  right  parallelopipedon  into  the  two  equal  tri- 
angular prisms,  aBd — e  and  Bed — g,  (Th.  1).  We  will 
now  compare  the  right  prism  with  the  oblique  triangular 
prism  on  the  same  side  of  the  diagonal  plane. 

The  volume  ABD  —  e  is  common  to  the  two  prisms, 
ABD — F  xnd  aBd — e  ;  and  the  volume  eFh — F,  which, 
added  to  this  common  part,  forms  the  oblique  triangular 
prism,  is  equal  to  the  volume  aBd — A,  which,  added  to 
the  common  part,  forms  the  right  triangular  prism.  For, 
since  ABFF  and  aBFe  are  parallelograms,  AF  =  ae,  and 
taking  away  the  common  part  Ae,  we  have  aA—eF;  and 
since  BFHD  and  BFhd  are  parallelograms,  we  have  DE 
=  dh ;  and  from  these  equals  taking  away  the  common 
part.  Dh,  we  have  dD  =  hE.    Now,  if  the  volume  eFh — H 


190  GEOMETRY. 

be  applied  to  the  volume  aBd — B,  the  base  eFh  falling 
on  the  equal  base  aBd,  the  edges  eE  and  hH  will  fall 
upon  aA  and  dB  respectively,  because  they  are  perpen- 
dicular to  the  base  aBd,  (Cor.  2,  Th.  3,  B.  VI),  and  the 
point  E  will  fall  upon  the  point  A,  and  the  point  H  upon 
the  point  B ;  hence  the  volume  eFh — H  exactly  coincides 
with  the  volume  aBd — B,  and  the  oblique  triangular 
prisrr.  ABB  —  E  is  equivalent  to  the  right  triangular 
prism  aBd — e. 

In  the  same  manner,  it  may  be  proved  that  the  oblique 
triangular  prism,  BQBOr,  is  equivalent  to  the  right  tri- 
angular prism,  Bcdg.  The  oblique  triangular  prism  on 
either  side  of  the  diagonal  plane  is,  therefore,  equivalent 
to  the  corresponding  right  triangular  prism ;  and,  as  the 
two  right  triangular  prisms  are  equal,  the  oblique  trian- 
gular prisms  are  equivalent. 

Hence  the  theorem ;  the  two  triangular  prisms,  etc. 

THEOREM    XI. 

The  volume  of  any  prism  whatever  is  measured  by  the  prod* 
net  of  the  area  of  its  base  and  altitude. 

For,  by  passing  planes  through  the  homologous  diag- 
onals of  the  upper  and  lower  bases  of  the  prism,  it  will 
be  divided  into  a  number  of  triangular  prisms,  each  of 
which  is  measured  by  the  product  of  the  area  of  its  base 
and  altitude.  Now,  as  these  triangular  prisms  all  have, 
for  their  common  altitude,  the  altitude  of  the  given 
prism,  when  we  add  the  measures  of  the  triangular 
prisms,  to  get  that  of  the  whole  prism,  we  shall  have, 
for  this  measure,  the  common  altitude  multiplied  by  the 
sum  of  the  areas  of  the  bases  of  the  triangular  prisms : 
that  is,  the  product  of  the  area  of  the  polygonal  base 
and  the  altitude  of  the  prism. 

Hence  the  theorem ;  the  volume  of  any  prism,  etc. 

Cor.  If  A  denote  the  area  of  the  base,  and  H  the  alti» 


BOOK    VII.  191 

hide  of  a  prism,  its  volume  will  be  expressed  by  A  X  11. 
Calling  this  volume  V,  we  have 

Denoting  by  A',  W,  and  V,  in  order,  the  area  of  the 
base,  altitude,  and  volume  of  another  prism,  we  have 

V  =  A1  x  Hf. 

Dividing  the  first  of  these  equations  by  the  second, 
u  ember  by  member,  we  have 

YL      A  x  H 

V  '  A'  x  R'y 
which  gives  the  proportion, 

V  :    V  : :  A  x  H  :  A1  x  H'. 
If  the  bases  are  equivalent,  this  proportion  becomes 

V  :  V  : :  H  :  H' ; 
anl  if  the  altitudes  are  equal,  it  reduces  to 

V  :   V  : :  A  :  A'. 
Hence,  prisms  of  equivalent  bases  are  to  each  other  as 
their  altitudes;  and  prisms  of  equal  altitudes  are  to  each  other 
as  their  bases. 

THEOREM    XII. 

A  plane  passed  through  a  pyramid  parallel  to  its  base, 
divides  its  edges  and  altitude  proportionally,  and  makes  a 
section,  which  is  a  polygon  similar  to  the  base. 

Let  ABODE — V  be  any  pyramid,  whose  base  is  in  the 
plane,  MN,  and  vertex  in  the  parallel  plane,  mn ;  and  let 
a  plane  be  passed  through  the  pyramid,  parallel  to  its 
base,  cutting  its  edges  at  the  points,  a  b,  c}  d,  e,  and  the 
altitude,  EF,  at  the  point  I.  By  joining  the  points,  a,  J, 
c,  etc.,  we  have  the  polygon  formed  by  the  intersection 
of  the  plane  and  the  sides  of  the  pyramid.  Eow,  we  are 
to  prove  that  the  edges,  VA,  VB,  etc.,  and  the  altitude, 
FE,  are  divided  proportionally  at  the  points,  a,  b,  etc., 
and  I;  and  that  the  polygon,  a,  b,  c,  d,  e,  is  similar  to  the 
base  of  the  pyramid. 


192 


GEOMETRY. 


bince  the  cutting  plane  is  parallel  to  the  base  of  the 
pyramid,  ab  is  parallel  to  AB,  (Th.  9,  B.  VI) ;  for  the 
same  reason,  bo  is  parallel  to  BO,  cd  to  OB,  etc.  Now, 
in  the  triangle  VAB,  because  ab  is  parallel  to  the  base 
AB,  we  have,  (Th.  17,  B.  II),  the  proportion, 

VA  :  Va  : :  VB  :  Vb. 

In  like  manner,  it  may  be  shown  that 

VB  :  Vb  : :  VO  :  Vc, 

and  so  on  for  the  other  lateral  edges  of  the  pyramid.  F 
being  the  point  in  which  the  perpendicular  from  F  pierces 
tbe  plane  mn,  and  I  the  point  in  which  the  parallel  secant 
plane  cuts  the  perpendicular,  if  we  join  the  points  F  and 
V,  and  also  the  points  I  and  e  by  straight  lines,  we  have 
in  the  triangle  FFV,  the  line  le  parallel  to  the  base  FV; 
hence  the  proportion 

VF  :  Ve  ::  FF  :  Fl 

Therefore,  the  plane  passed  through  the  pyramid  par- 
allel to  its  base,  divides  the  altitude  into  parts  which  have 


BOOK    VII.  193 

to  each   other  the  same  ratio  as  the  parts  into  which  it 
divides  the  edges. 

Again,  since  ab  is  parallel  to  AB,  and  be  to  BO,  the 
angle  abc  is  equal  to  the  angle  ABC,  (Th.  17,  B.  VI.) ;  in 
the  same  manner  we  may  show  that  each  angle  in  the 
polygon,  abode,  is  equal  to  the  corresponding  angle  in  the 
polygon,  ABODE-,  therefore  these  polygons  are  mutually 
equiangular.  But,  because  the  triangles  VBA  and  Vba 
are  similar,  their  homologous  sides  give  the  proportion 

Vb  :  VB  ::  ah  :  AB; 
and  because  the  triangles  Vbc  and  VBO  are  similar,  we 
also  have  the  proportion 

Vb  :  VB  ::  be  :  BO. 

Since  the  first  couplets  in  these  two  proportions  are  the 
same,  the  second  couplets  are  proportional,  and  give 
ab  :  AB  : :  be  :  BO. 

By  a  like  process,  we  can  prove  that 
be  :  BO  ::  ed  :  OD, 

and  that  cd  :  CD  ::  de  :  DE, 

and  so  on,  for  the  other  homologous  sides  of  the  two 
polygons. 

Hence,  the  two  polygons  are  not  only  mutually  equi- 
angular, but  the  sides  about  the  equal  angles  taken  in  the 
same  order  are  proportional,  and  the  polygons  are  there- 
fore similar,  (Def.  16,  B.  II). 

Hence  the  theorem;  a  plane  passed  through  a  pyramid, 
etc. 

Cor,  1.  Since  the  areas  of  similar  polygons  are  to  each 
other  as  the  squares  of  their  homologous  sides,  (Th.  22, 
B.  H),  we  have 

area  abode  :  area  ABODE  i  ab*  i  AB*. 

But,         ab  :  AB  ::  Va  :  VA   ::  Fl  :  FE; 

hence,  ab*  :  AB2  ::  jf  :  FE*: 

therefore,      area  abode  :  area  ABODE  :  Fl*  :  FE*. 
17  N 


194  GEOMETRY. 

That  is,  the  area  of  a  section  parallel  to  the  base  of  a 
pyramid,  is  to  the  area  of  the  base,  as  the  square  of  the 
perpendicular  distance  from  the  vertex  of  the  pyramid  to 
the  section,  is  to  the  square  of  the  altitude  of  the  pyramid. 

Cor.  2.  Let  V— ABODE  and  X—RST  be  two  pyra- 
mids, having  their  bases  in  the  plane  MN,  and  their  ver- 
tices in  the  parallel  plane  mn  ;  and  suppose  a  plane  to  be 
passed  through  the  two  pyramids  parallel  to  the  common 
plane  of  their  bases,  making  in  the  one  the  section  abode, 
and  in  the  other  the  section  rsL 

Kow,area ABODE:  area  abode :  :AB2 :  ab\  (Th.22,B.II), 

and      "  EST:    "        rst  ::~M2  :r~s~\ 

But,  AB  :  ab  : :    VB  :  Vb, 

and  BS  :  rs  : :  XR  :  Xr. 

Because  the  plane  which  makes  the  sections  is  parallel 
to  the  planes  MN  and  mn,  we  have,  (Th.  11,  B.  YI), 
VB  :  Vb  ::  XR  :  Xr; 
therefore,  (Cor.  2,  Th.  6,  B.  II),  AB  :ab::RS:  rs. 
By  squaring,       AB2  :  ab2  :  RS2  :  rs2; 
hence,   area  ABODE :  area  abode  : :  area  RST :  area* rst. 

That  is,  if  two  pyramids  having  equal  altitudes,  and  their 
bases  in  the  same  plane,  be  cut  by  a  plane  parallel  to  the  com* 
mon  plane  of  their  bases,  the  areas  of  the  sections  will  be 
proportional  to  the  areas  of  the  bases ;  and  if  the  bases  are 
equivalent,  the  sections  will  also  be  equivalent. 

THEOREM    XIII. 

If  two  triangular  pyramids  have  equivalent  bases  and 
equal  altitudes,  they  are  equal  in  volume. 

Let  V—  ABO  and  v—abc  be  two  triangular  pyramids, 
having  the  equivalent  bases,  ABO  and  abc,  and  let  the 
altitude  of  each  be  equal  to  OX;  then  will  these  two 
pyramids  be  equivalent. 


BOOK    VII 


195 


-i  i 

t/ 

AFT"'? 

1  //  y 

1/      -i'H 

m 

//  /A /I 

F 

~^f 

^B 

Place  tlie  bases  of  the  pyramids  on  the  same  plane, 
with  their  vertices  in  the  same  direction,  and  divide  the 
altitude  into  any  number  of  equal  parts.  Through  the 
points  of  division  pass  planes^  parallel  to  the  plane  of  the 
bases ;  the  corresponding  sections  made  in  the  pyramids 
by  these  planes  are  equivalent,  (Th.  12,  Cor.  2) ;  that  is, 
the  triangle  DJEF  is  equivalent  to  the  triangle  def,  the 
triangle  Or  HI  to  the  triangle  glii,  etc. 

JSTow,  let  triangular  prisms  be  constructed  on  the  tri- 
angles ABO,  DEF,  etc.,  of  the  pyramid  V—  ABO,  these 
prisms  having  their  lateral  edges  parallel  to  the  edge, 
VO,  of  the  pyramid,  and  the  equal  parts  of  the  altitude, 
OX,  for  their  altitudes.  Portions  of  these  prisms  will  be 
exterior  to  the  pyramid  V — ABO,  and  the  sum  of  their 
volumes  will  exceed  the  volume  of  the  pyramid. 

On  the  bases  clef,  ghi,  etc.,  in  the  other  pyramid,  con- 
struct interior  prisms,  as  represented  in  the  figure, 
their  lateral  edges  being  parallel  to  vc,  and  their  alti- 
tudes also  the  equal  parts  of  the  altitude,  OX.  Portions 
of  the  pyramid,  v — dbc,  will  be  exterior  to  these  prisms, 


196  GEOMETRY. 

and  the  volume  of  the  pyramid  will  exceed  the  sum  of 
the  volumes  of  the  prisms. 

Since  the  sum  of  the  exterior  prisms,  constructed  in 
connection  with  the  pyramid  V — ABO,  is  greater  than 
the  pyramid,  and  the  sum  of  the  interior  prisms,  con- 
structed in  connection  with  the  pyramid  v — abc,  is  less 
than  this  pyramid,  it  follows  that  the  difference  of  these 
eums  is  greater  than  the  difference  of  the  pyramids  them- 
selves. But  the  second  exterior  prism,  or  that  on  the 
base  DEF,  is  equivalent  to  the  first  interior  prism,  or 
that  on  the  base  def,  and  the  third  exterior  prism  is 
equivalent  to  the  second  interior  prism,  (Th.  10,  Cor.  2), 
and  so  on.  That  is,  beginning  with  the  second  prism  from 
the  base  of  the  pyramid,  V — ABO,  and  taking  these 
prisms  in  order  towards  the  vertex  of  the  pyramid,  and 
comparing  them  with  the  prisms  in  the  pyramid,  v — abc, 
beginning  with  the  lowest,  and  taking  them  in  order 
toward  the  vertex  of  this  pyramid,  we  find  that  to  each 
exterior  prism  of  the  pyramid,  V — ABO,  exclusive  of 
the  first  or  lowest,  there  is  a  corresponding  equivalent 
interior  prism  in  the  pyramid,  v — abc. 

Hence  the  prism,  ABODFF,  is  the  difference  between 
the  sum  of  the  prisms  constructed  in  connection  with 
the  pyramid,  V— ABO,  and  the  sum  of  the  interior 
prisms  constructed  in  the  pyramid,  v — abc.  But  the,  first 
sum  being  a  volume  greater  than  the  pyramid,  V—  ABO, 
and  the  second  sum  a  volume  less  than  the  pyramid, 
v — abc,  it  follows  that  the  volumes  of  the  pyramids  differ 
by  less  than  the  prism,  ABOBEF. 

Now,  however  great  the  number  of  equal  parts  into 
which  the  altitude,  OX,  be  divided,  and  the  correspond- 
ing number  of  prisms  constructed  in  connection  with 
each  pyramid,  it  would  still  be  true  that  the  difference 
between  the  volumes  of  the  pyramids  would  be  less  than 
the  volume  of  the  lowest  prism  of  the  pyramid  V—  ABO; 
Dutwhenwe  make  the  number  of  equal  parts  into  which 


BOOK   VII.  19T 

the  altitude  is  divided  indefinitely  great,  the  vok  me  ot 
this  prism  becomes  indefinitely  small :  that  is,  the  differ- 
ence between  the  volumes  of  the  pyramids  is  less  thau 
an  indefinitely  small  volume ;  or,  in  other  words,  there 
is  no  assignable  difference  between  the  two  pyramids, 
and  they  are,  therefore,  equivalent. 

Ilence  the  theorem ;  if  two  triangular  pyramids,  etc. 

THEOREM    XIV. 

Any  triangular  pyramid  is  one  third  of  the  triangul  xr 
prism  having  the  same  base  and  equal  altitude. 

Let  F — ABO  be  a  triangular  pyramid,  and  througa  F 
pass  a  plane  parallel  to  the  plane  of  the  base,  ABO.    Iu 
this  plane,  through  F,  construct  the 
triangle,  FDE,  having  its  sides,  FD,  E 

DE,  and  EF,  parallel  and  equal  to  BO,  /^T       ~~7\ 

OA,  and  AB,  respectively.     The  tri-  /     0>\      / 

angle,  FDE,   may  be  taken  as   the         /    /h\ 
upper  base  of  a  triangular  prism  of  'I   'V 

which  the  lower  base  is  ABO.  "\.  ~\ / 

Now,  this  triangular  prism  is  com-  b 

posed  of  the  given  triangular  pyramid, 
F — ABO,  and  of  the  quadrangular  p/ramid,  F — A  ODE. 
This  last  pyramid  may  be  divided  by  a  plane  through  the 
three  points,  0,  F,  and  F,  into  tha  two  triangular  pyra- 
mids, F—DEO  and  F—AOE.  But  the  pyramid,  J7— 
DEO,  may  be  regarded  as  h^ing  the  triangle,  EFD, 
equal  to  the  triangle,  ABO,  for  its  base,  and  the  point,  0, 
for  its  vertex.  The  two  pyramids,  F — ABO  and  0 — DEF, 
have  equal  bases  and  equal  altitudes ;  they  are  therefore 
equivalent,  (Th.  13).  Again,  the  two  pyramids,  F — DEO 
and  F — AOE}  have  a  common  vertex,  and  equivalent  bases 
in  the  same  plane,  and  they  are  also  equivalent.  There- 
fore, the  triangular  prism,  ABODEF,  is  composed  of 

n* 


198 


GEOMETRY. 


three  *tpiv«dent  triangular  pyramids,  one  of  which  is  the 
given  triangular  pyramid,  F — ABC. 

Hetice  the  theorem ;  any  triangular  pyramid  is  one  third 
of  the  triangular  prism,  etc. 

Cor.  The  volume  of  the  triangular  prism  heing  meas- 
ured hy  the  product  of  its  hase  and  altitude,  the  volume  of 
a  triangular  pyramid  is  measured  by  one  third  of  the  product 
of  its  base  and  altitude. 

THEOREM    XV. 

The  volume  of  any  pyramid  whatever  is  measured  by  one 
third  of  the  product  of  its  base  and  altitude. 

Let  V — ABCDE  be  any  pyramid ;  then  will  its  volume 
be  measured  by  one  third  of  the  product  of  its  base  and 
altitude. 

In  the  base  of  the  pyramid,  draw  the 
diagonals,  AD  and  A  C,  and  through 
its  vertex  and  these  diagonals,  pass 
planes,  thus  dividing  the  pyramid  into 
a  number  of  triangular  pyramids 
having  the  common  vertex  F",  and  the 
altitude  of  the  given  pyramid  for  their 
common  altitude. 

Now,  each  of  these  triangular  pyra- 
mids is  measured  by  one  third  of 
the  product  of  its  base  and  altitude, 
(Cor.,  Th.  14),  and  their  sum,  which 
constitutes  the  polygonal  pyramid,  is 
therefore  measured  by  one  third  of 
the  product  of  the  sum  of  the  trian- 
gular bases  and  the  common  altitude ;  but  the  sum  of  the 
triangular  bases  constitutes  the  polygonal  base,  ABCDE. 

Hence  the  theorem ;  the  volume  of  any  pyramid  what- 
ever, etc. 

Cor.  1.  Denote,  by  B,  ff,  and  V,  respectively,  the  base, 
altitude,  and  volume  of  one  pyramid,  and  by  B'f  IF,  and 


BOOK   VII.  199 

P,  the  base,  altitude,  and  volume  of  another ;  then  we 
Bhall  have 

and  V  =  i'Bf  x  W. 

Dividing  the  first  of  these  equations  by  the  second, 
member  by  member,  we  have 

V  _B  x  E 

V  Bf  x  E'9 
which,  in  the  form  of  a  proportion,  gives 

V  :    V  : :  B  X  E  :  B'  x  E>. 

From  this  proportion  we  deduce  the  following  conse- 
quences : 

1st.  Pyramids  are  to  each  other  as  the  products  of  their 
bases  and  altitudes. 

2d.  Pyramids  having  equivalent  bases  are  to  each  other  a* 
their  altitudes. 

3d.  Pyramids  having  equal  altitudes  are  to  each  other  as 
their  bases. 

Cor.  2.  Since  a  prism  is  measured  by  the  product  of 
its  base  and  altitude,  and  a  pyramid  by  one  third  of  the 
product  of  its  base  and  altitude,  we  conclude  that  any 
pyramid  is  one  third  of  a  prism  having  an  equivalent  base  and 
equal  altitude 

THEOREM    XVI. 

The  volume  of  the  frustum  of  a  pyramid  is  equivalent  to 
the  sum  of  the  volumes  of  three  pyramids,  each  of  which  has 
an  altitude  equal  to  that  of  the  frustum,  and  whose  bases  are, 
respectively,  the  lower  base  of  the,  frustum,  the  upper  base  of 
the  frustum,  and  a  mean  proportional  between  these  bases. 

Let  V— ABODE  and  X—RST  be  two  pyramids,  the 
one  polygonal  and  the  other  triangular,  having  equiva- 
lent bases  and  equal  altitudes ;  and  let  their  bases  be 
placed  on  the  plane  MN,  their  vertices  falling  on  the 
parallel  plane  ran.     Pass  through  the  pyramids  a  plane 


200 


GEOMETRY. 


parallel  to  the  common  plane  of  their  bases,  cutting  o  .it 
the  sections  abode  and  rst ;  these  sections  are  equivalent, 
(Th.  12,  Cor.  2),  and  the  pyramids,  V— abode  and  X—rst, 
are  equivalent,  (Th.  13).  Now,  since  the  pyramids, 
V— ABODE  and  X—BST,  are  equivalent,  if  from  the 
first  we  take  the  pyramid,  V — abode,  and  from  the  second, 
the  pyramid,  X — rst,  the  remainders,  or  the  frusta, 
ABODE— a  and  BST—r,  will  be  equivalent. 

If,  then,  we  prove  the  theorem  in  the  case  of  the  frus- 
tum of  a  triangular  pyramid,  it  will  be  proved  for  the 
frustum  of  any  pyramid  whatever. 

Let  ABO — D  be  the  frustum  of  a 
triangular  pyramid.  Through  the 
points  D,  B,  and  0,  pass  a  plane, 
and  through  the  points  D,  O,  and 
E,  pass  another,  thus  dividing  the 
frustum  into  three  triangular  pyra- 
mids, viz.,  D—ABO,  O—DEF,  and 
D—BEO. 

Now,  the  first  of  these  has,  for  its 


BOOK   VII.  201 

base,  the  lower  base  of  the  frustum,  and  for  its  altitude 
the  altitude  of  the  frustum,  since  its  vertex  is  in  the 
upper  base ;  the  second  has,  for  its  base,  the  upper  base 
of  the  frustum,  and  for  its  altitude  the  altitude  of  the 
frustum,  since  its  vertex  is  in  the  lower  base.  Hence, 
these  are  two  of  the  three  pyramids  required  by  the 
enunciation  of  the  theorem ;  and  we  have  now  only  to 
prove  that  the  third  is  equivalent  to  one  having,  for  its 
basa,  a  mean  proportional  between  the  bases  of  the  frus- 
tum, and  an  altitude  equal  to  that  of  the  frustum. 

In  the  face  ABED,  draw  HB  parallel  to  BE,  and 
draw  HE  and  HO.  The  two  pyramids,  D — BEO  and 
H—BEO,  are  equivalent,  since  they  have  a  common 
base  and  equal  altitudes,  their  vertices  being  in  the  line 
BH,  which  is  parallel  to  the  plane  of  their  common 
base,  (Th.  7,  B.  VI).  "We  may,  therefore,  substitute  the 
pyramid,  H—BEO,  for  the  pyramid,  D—BEC.  But  the 
triangle,  BOH,  may  be  taken  as  the  base,  and  E  as  the 
vertex  of  this  new  pyramid ;  hence,  it  has  the  required 
altitude,  and  we  must  now  prove  that  it  has  the  required 
base. 

The  triangles,  ABO  and  HBO,  have  a  common  vertex, 
and  their  bases  in  the  same  line ;  hence,  (Th.  16,  B.  II), 

A  ABO  :  A  HBO  n  AB  :  HB  ::  AB  :  BE.    (1) 

In  the  triangles,  BEE  and  HBO,  \_E^\_B,  and 
BE=HB;  hence,  if  BEE  be  applied  to  HBO,  L  E  fall- 
ing on  L  B,  an(i  tne  side  BE  on  HB,  the  point  B  will 
fall  on  H,  and  the  triangles,  in  this  position,  will  have  a 
common  vertex,  H,  and  their  bases  in  the  same  line ; 
hence, 

A  HBO  :  A  BEE  : :  BO  :  EF.     (2) 

But,  because  the  triangles,  ABO  and  BEE,  are  similar, 

we  nave 

AB  .  BE  ii  BO  i  EF.     (3) 

From  proportions  (1),  (2),  and  (3),  we  have,  (Th.  6, 

b  n), 


202  GEOMETRY. 

A  ABO  :  A  HBO  : :  A  HBO  :  A  DEF; 
that  is,  the  base,  HBO,  is  a  mean  proportional  between 
the  lower  and  upper  bases  of  the  frustum. 

Hence  the  theorem ;  the  volume  of  the  frustum  of  a  pyra- 
mid, etc, 

THEOREM    XVII. 

The  convex  surface  of  any  right  'pyramid  is  measured  by 
the  perimeter  of  its  base,  multiplied  by  one  half  its  slant  height. 

Let  S—ABOBEF  be  a  right  pyramid,  s 

of  which  SH  is  the  slant  height ;  then  will  k 

its  convex  surface  have,  for  its  measure,  /M 

±SH(AB  +  BO+OB  +  BE+EF+FA).  Mm 

Since  the  base  is  a  regular  polygon,  and       /e/LjJa\ 
the  perpendicular,  drawn  to  its  plane  from    ^1/  I  \     \\ 
S,  passes  through  its  center,  the   edges,      \  /  i    / 
SA,  SB,  SO,  etc.,  are  equal,  (Th.  4,  B.  YI),        ahb 
and  the  triangles  SAB,  SBC,  etc.,  are  equal,  and  isosceles, 
each  having  an  altitude  equal  to  SH. 

Now,  AB  x  \SH  measures  the  area  of  the  triangle, 
SAB ;  and  BO  X  %SH  measures  the  area  of  the  triangle, 
SBO;  and  so  on,  for  the  other  triangular  faces  of  the 
pyramid.  By  the  addition  of  these  different  measures, 
we  get 

%SH(AB  +  BO +  OB  +  BE  +  EF +  FA), 
as  the  measure  of  the  total  convex  surface  of  the  pyramid. 

Hence  the  theorem ;  the  convex  surface  of  any  right 
pyramid,  etc, 

THEOREM  XVIII. 

The  convex  surface  of  the  frustum  of  any  right  pyramid  is 
measured  by  the  sum  of  the  perimeters  of  the  two  bases,  mul- 
tiplied by  one  half  the  slant  height  of  the  frustum. 

Let  ABOBEF — d  be  the  frustum  of  a  right  pyramid; 
then  will  its  convex  surface  be  measured  by 

$Ilh(AB-+  BC4-CD+DE-\  EF+FA+ab+bc+cd+de+ef+fa), 


book  vir 


203 


For,  the  upper  base,  abcdef,  of  the 
frustum  is  a  section  of  a  pyramid 
by  a  plane  parallel  to  the  lower 
base,  (Def.  14),  and  is,  therefore, 
similar  to  the  lower  base,  (Th.  12). 
But  the  lower  base  is  a  regular 
polygon,  (Def.  12);  hence,  the  up- 
per base  is  also  a  regular  polygon, 
of  the  same  name;  and  as  ab  and 
AB  are  intersections  of  a  face  of 
the  pyramid  by  two  parallel  planes,  A     H     B 

they  are  parallel.  For  the  same  reason,  be  is  parallel  to 
BO,  cd  to  OB,  etc.,  and  the  lateral  faces  of  the  frustum 
are  all  equal  trapezoids,  each  having  an  altitude  equal 
to  Hh,  the  slant  height  of  the  frustum. 

The  trapezoid  ABba  has,  for  its  measure,  \Hh(AB+ab), 
(Th.  34,  Book  I) ;  the  trapezoid  BOcb  has,  for  its  meas- 
ure, \Hh{BO +  be),  and  so  on,  for  the  other  lateral  faces 
of  the  frustum. 

Adding  all  these  measures,  we  find,  for  their  sum, 
which  is  the  whole  convex  surface  of  the  frustum, 

IHh  {AB+  B  C+  CD+DE+EF+  FA+ab+bc+cd+de+ef+fa). 

Hence  the  theorem ;  the  convex  surface  of  the  frustum. 


THEOREM    XIX. 

The  volumes  of  similar  triangular  prisms  are  to  each  other 
as  the  cubes  constructed  on  their  homologous  edges. 

Let  ABO—  F  ana 

abc— /be  two  similar 
triangular  prisms ; 
then  will  their  vol- 
umes be  to  each 
other  as  the  cubes, 
whose  edges  are  the 
homologous    edges 


204  GEOMETRY. 

AB  and  a b,  or  as  the  cubes,  whose  edges  are  the  homol- 
ogous edges  BE  and  be,  etc.  Since  the  prisms  are  similar, 
the  solid  angles,  whose  vertices  are  B  and  b,  are  equal ; 
and  the  smaller  prism,  when  so  applied  to  the  larger  that 
these  solid  angles  coincide,  will  take,  within  the  larger, 
the  position  represented  by  the  dotted  lines.  In  this 
position  of  the  prisms,  draw  EH  perpendicular  to  the 
plane  of  the  base  ABO,  and  join  the  foot  of  the  perpen- 
dicular to  the  point  B,  and  in  the  triangle  BEIT  draw, 
through  e,  the  line  eh,  parallel  to  EH',  then  will  EH 
represent  the  altitude  of  the  larger  prism,  and  eh  that  of 
the  smaller. 

Now,  as  the  bases  ABO  and  aBc,  are  homologous  faces, 
they  are  similar,  and  we  have,  (Th.  20,  Book  II), 

A  ABO  :  A  aBc  : :  AB*  CoE*         ( 1 ) 

But  the  A's  BEH  and  Beh  are  equiangular,  and  there 
fore  similar,  and  their  homologous  sides  give  the  propor- 
tion 

BE  :  Be  ::  EH  :  eh         (2) 

and  from  the  homologous  sides  of  the  similar  faces, 
ABED  and  aBed,  we  also  have 

BE  :  Be  ::  AB  :  aB         (3) 

Proportions  (2)  and  (3 ),  having  an  antecedent  and  con 
sequent  the  same  in  both,  we  have,  (Th.  6,  B.  II), 

EH  :  eh  w  AB  \  aB         (4) 

By  the  multiplication  of  proportions  (1)  and  (•*),  term 
Dy  term,  we  get 

A  ABO  X  EH :  A  aBc  X  eu  : :  AB* :  aB* 

But  A  ABO  X  EH  measures  the  volume  of  the  larger 
prism,  and  A  aBc  x  eh  measures  the  volume  of  the 
smaller. 

Hence  the  theorem;  the  volumes  of  similar  triangular 
prisms^  etc. 


BOOK    VII.  205 

Cor.  1.  The  volumes  of  two  similar  prisms  having  any 
bases  whatever,  are  to  each  other  as  the  cubes  constructed  on 
their  homologous  edges. 

For,  if  planes  be  passed  through  any  one  of  the  late<-al 
edges,  and  the  several  diagonal  edges,  of  one  of  these 
prisms,  this  prism  will  be  divided  into  a  number  of  smaller 
triangular  prisms.  Taking  the  homologous  edge  of  the 
other  prism,  and  passing  planes  through  it  and  the  seve- 
ral diagonal  edges,  this  prism  will  also  be  divided  into 
the  same  number  of  smaller  triangular  prisms,  similar  to 
those  of  the  first,  each  to  each,  and  similarly  placed. 

]STow,  the  similar  smaller  prisms,  being  triangular,  are 
to  each  other  as  the  cubes  of  their  homologous  edges ; 
and  being  like  parts  of  the  larger  prisms,  it  follows  that 
the  larger  prisms  are  to  each  other  as  the  cubes  of  the 
homologous  edges  of  any  two  similar  smaller  prisms.  But 
the  homologous  edges  of  the  similar  smaller  prisms  are 
to  each  other  as  the  homologous  edges  of  the  given 
prisms ;  hence  we  conclude  that  the  given  prisms  are  to 
each  other  as  the  cubes  of  their  homologous  edges. 

Cor.  2.  The  volumes  of  two  similar  pyramids  having  any 
bases  whatever,  are  to  each  other  as  the  cubes  constructed  on 
their  homologous  edges. 

For,  since  the  pyramids  are  similar,  their  bases  are 
similar  polygons ;  and  upon  them,  as  bases,  two  similar 
prisms  may  be  constructed,  having  for  their  altitudes,  the 
altitudes  of  their  respective  pyramids,  and  their  lateral 
edges  parallel  to  any  two  homologous  lateral  edges  of  the 
pyramids. 

Now,  these  similar  prisms  are  to  each  other  as  the  cubes 
of  their  homologous  edges,  which  may  be  taken  as  the 
homologous  sides  of  their  bases,  or  as  their  lateral  edges, 
which  were  taken  equal  and  parallel  to  any  two  arbitrarily 
assumed  homologous  lateral  edges  of  the  two  pyramids  ; 
hence  the  pyramids  which  are  thirds  of  their  respective 
prisms,  are  to  each  other  as  the  cubes  constructed  on  any 
two  homologous  edges. 

18 


206 


GEOMETRY. 


Cor.  3.  The  volumes  of  any  two  similar  polyedrons  arz  U 
each  other  as  the  cubes  constructed  on  their  homologous  edges. 

"For,  by  passing  planes  through,  the  vertices  of  the 
homologous  solid  angles  of  such  polyedrons,  they  may 
both  be  divided  into  the  same  number  of  triangular 
pyramids,  those  of  the  one  similar  to  those  of  the  other, 
each  to  each,  and  similarly  placed. 

Now,  any  two  of  these  similar  triangular  pyramids  are 
to  each  other  as  the  cubes  of  their  homologous  edges ; 
and  being  like  parts  of  their  respective  polyedrons,  it 
follows  that  the  polyedrons  are  to  each  other  as  the  cubes 
of  the  homologous  edges  of  any  two  of  the  similar  tri- 
angular pyramids  into  which  they  may  be  divided.  But 
the  homologous  edges  of  the  similar  triangular  pyramids 
are  to  each  other  as  the  homologous  edges  of  the  poly- 
edrons ;  hence  the  polyedrons  are  to  each  other  as  the 
cubes  of  their  homologous  edges. 


THEOREM    XX. 


The  convex  surface  of  the  frustum  of  a  cone  is  measured 
by  the  product  of  the  slant  height  and  one  half  the  sum  of 
the  circumferences  of  the  bases  of  the  frustum. 

Let  ABCD — abed  be  the  frustum  of 

a  cone ;  then  will  its  convex  surface  be 

,  ,       .         (circ.  00  +  circ.  oc) 
measured  by  Aa  x  i — '-, 

in  which  the  expression,  circ.  00,  de- 
notes the  circumference  of  the  circle 
of  which  00  is  the  radius.  Inscribe  in 
the  lower  base  of  the  frustum,  a  regu- 
lar polygon  having  any  number  of 
sides,  and  in  the  upper  base  a  similar 
polygon,  having  its  sides  parallel  to 
those  of  the  polygon  in  the  lower  base. 


These  polygons 


BOOK   VII.  207 

may  be  taken  as  the  bases  of  the  Irustum  of  a  right 
pyramid  inscribed  in  the  frustum  of  the  cone. 

Now,  however  great  the  number  of  sides  of  the  in- 
scribed polygons,  the  convex  surface  of  the  frustum  of 
the  pyramid  is  measured  by  its  slant  height  multiplied  by 
one  half  the  sum  of  the  perimeters  of  its  two  bases, 
(Th.  18) ;  but  when  we  reach  the  limit,  by  making  the 
number  of  sides  of  the  polygon  indefinitely  great,  the 
slant  height,  perimeters  of  the  bases,  and  convex  surfaco 
of  the  frustum  of  the  pyramid  become,  severally,  the 
slant  height,  circumferences  of  the  bases,  and  convex  sur- 
face of  the  frustum  of  the  cone. 

Hence  the  theorem ;  the  convex  surface  of  the  frustum, 
etc. 

Cor.  1.  If  we  make  oc  =  OC,  and,  consequently,  circ. 
oc  =  circ.  OC,  the  frustum  of  the  cone  becomes,  a  cylin- 
der, and  the  half  sum  of  the  circumferences  of  the  bases 
becomes  the  circumference  of  either  base  of  the  cylinder, 
and  the  slant  height  of  the  frustum,  the  altitude  of  the 
cylinder.  Hence,  the  convex  surface  of  a  cylinder  is  meas- 
ured by  the  circumference  of  the  base  multiplied  by  the  alti- 
tude of  the  cylinder. 

Cor.  2.  If  we  make  oc  =  0,  the  frustum  of  the  cone 
becomes  a  cone.  Hence,  the  convex  surface  of  a  cone  is 
measured  by  the  circumference  of  the  base  multiplied  by  one 
half  the  slant  height  of  the  cone. 

Cor.  3.  If  through  E,  the  middle  point  of  Co,  the  line 
Ff  be  drawn  parallel  to  Oo,  and  Em  perpendicular  to 
Oo,  the  line  oc  being  produced,  to  meet  Ff  at/,  we  have, 
because  the  A's  EFC  and  Efc  are  equal, 

Em  =   O0  +  M. 

2S 

if  we  multiply  both  members  of  this  equation  by  2*, 
we  have 


208  GEOMETRY. 

that  is,  circ.  Em  is  equal  to  one  half  the  sum  of  the  cir 
cumferences  of  the  two  hases  of  the  frustum.  Hence,  the 
convex  surface  of  the  frustum  of  a  cone  is  measured  by  the. 
circumference  of  the  section  made  by  a  plane  half  way  between 
the  two  bases,  and  parallel  to  them,  multiplied  by  the  slant 
height  of  the  frustum. 

Cor.  4.  If  the  trapezoid,  OCeo,  he  revolved  ahout  Oo 
as  an  axis,  the  inclined  side,  Cc,  will  generate  the  con- 
vex surface  of  the  frustum  of  a  cone,  of  which  the  slant 
height  is  Cc,  and  the  circumferences  of  the  hases  are  circ. 
OC  and  circ.  oc.  Hence,  if  a  trapezoid,  one  of  whose  sides 
is  perpendicular  to  the  two  parallel  sides,  be  revolved  about 
the  perpendicular  side  as  an  axis,  it  will  generate  the  frustum 
of  a  cone,  the  inclined  side  opposite  the  axis  generating  tho 
convex  surface,  and  the  parallel  sides  the  bases  of  the  frustum. 

THEOREM    XXI. 

The  volume  of  a  cone  is  measured  by  the  area  of  its  base 
multiplied  by  one  third  of  its  altitude. 

Let  V— ABC,  etc.,  he  a  cone;  then 
will  its  volume  he  measured  hy  area 
ABC,  etc.,  multiplied  hy  iVO. 

Inscribe,  in  the  base  of  the  cone,  any 
regular  polygon,  as  ABCDEF,  which 
may  he  taken  as  the  base  of  a  right  pyra- 
mid, of  which  V  is  the  vertex.  The 
volume  of  this  inscribed  pyramid  will 
have,  for  its  measure,  (Th.  15), 

polygon  ABCDEF  x  iVO. 

Now,  however  great  the  number  of  sides  of  the  pol  f- 
gon  inscribed  in  the  base  of  the  cone,  it  will  still  ho  d 
true  that  the  pyramid  of  which  it  is  the  base,  and  who-e 
vertex  is  V,  will  be  measured  by  the  area  of  the  poly- 
gon, multiplied  by  one  third  of  VO;  but  when  we 
reach  the  limit,  by  making  the  number  of  sides  indefi- 


BOOK   VII.  209 

uitely  great,  the  polygon  becomes  the  ^cle  in  which  it 
is  inscribed,  and  the  pyramid  become   +he  cone. 

Hence  the  theorem ;  the  volume  of  a  cone,  etc. 

Cor.  1.  If  R  denote  the  radius  of  the  base  of  a  cone, 
and  ff  its  altitude,  or  axis,  its  volume  will  be  expressed 

by 

iff  x   *i22; 

hence,  if  V  and  V  designate  the  volume ~>  of  two  cones, 
of  which  R  and  R'  are  the  radii  of  the  bases,  and  H  and 
ff'  the  altitudes,  we  have 

V:  V  ::  J#x  *R2 :  \H'  x  «R'7 ;:  Bx«&  :  ff'  X  «R'\ 

From  this  proportion  we  conclude, 

First.  That  cones  having  equal  altitudes  are  to  each  other 
as  their  bases. 

Second.  That  cones  having  equal  bases  are  to  each  other 
as  their  altitudes. 

Cor.  2.  Retaining  the  notation  above,  we  have 

YL      El     ^1    m 

V    "    H  x    R*'     l   J 
and,  if  the  two  cones  are  similar, 

ff  :  ff!  ::  R  :  i2'; 
ff1        R'    ,  ff"        R'* 

or>        ~ff  =  R'>hence>ir  =  -&' 

By  substituting  for  the  factors,  in  the  second  member 
of  eq.  ( 1 ),  their  values  successively,  and  resolving  into  a 
proportion,  we  get 

V  :   V  ::  R*  :  R"; 
and  V  :   V  ::  ff3  :  H". 

Hence,  similar  cones  are  to  each  other  as  the  cubes  of  the 
radii  of  their  bases,  and  also  as  the  cubes  of  their  altitudes. 

Cor.  3.  A  cone  is  equivalent  to  a  pyramid  having  an  equiv> 
alent  base  and  an  equal  altitude. 
18*  0 


210 


GEOMETRY. 


THEOREM    XXII. 

'The  volume  of  the  frustum  of  a  cone  is  equivalent  to  the 
sum  of  the  volumes  of  three  cones,  having  for  their  common 
altitude  the  altitude  of  the  frustum,  and  for  their  several 
bases,  the  bases  of  the  frustum  and  a  mean  proportional  be- 
tween them. 

Let  ABQD — abed  be  the  frustum  of  a 
cone ;  then  will  its  volume  be  equiva- 
lent to  the  sum  of  the  volumes,  having 
Oo  for  their  common  altitude,  and  for 
their  bases,  the  circles  of  which,  OO,  oc, 
and  a  mean  proportional  between  OO 
and  oc,  are  the  respective  radii. 

Inscribe  in  the  lower  base  of  the  frus- 
tum any  regular  polygon,  and  in  the 
upper  base  a  similar  polygon,  having 
its  sides  parallel  to  those  of  the  first.  These  polygons 
may  be  taken  as  the  bases  of  the  frustum  of  a  right  pyra- 
mid inscribed  in  the  frustum  of  the  cone. 

The  volume  of  the  frustum  of  the  pyramid  is  equiva- 
lent to  the  sum  of  the  volumes  of  three  pyramids,  having 
for  their  common  altitude  the  altitude  of  the  frustum, 
and  for  their  several  bases  the  bases  of  the  frustum,  and 
a  mean  proportional  between  them,  (Th.  16). 

Eow,  however  great  the  number  of  sides  of  the  poly- 
gons inscribed  in  the  bases  of  the  frustum  of  the  cone, 
this  measure  for  the  volume  of  the  frustum  of  the  pyr* 
mid,  of  which  they  are  the  bases,  still  holds  true ;  biu 
when  we  reach  the  limit,  by  making  the  number  of  the 
sides  of  the  polygon  indefinitely  great,  the  polygons  be- 
come the  circles,  the  frustum  of  the  pyramid  becomes 
the  frustum  of  the  cone,  and  the  three  partial  pyramids, 
whose  sum  is  equivalent  to  the  frustum  of  the  pyramid, 
become  three  partial  cones,  whose  sum  is  equivalent  to 
the  frustum  of  the  cone. 


BOOK    VII. 


211 


Hence  the  theorem ;  the  volume  of  the  frustum  of  a  cone,  etc. 

Cor.  1.  Let  R  denote  the  radius  of  the  lower  base,  Rf 
that  of  the  upper  base,  and  IT  the  altitude  of  the  frustum 
of  a  cone;  then  will  its  volume  be  measured,  (Th.  21),  by 

IE  x  *i22  +  iH  x  «Rn  +  iffx  *R  x  R', 
since  *R  x  R'  expresses  the  area  of  a  circle  which  is  a 
mean  proportional  between  the  two  circles,  whose  radii 
arc  R  andi2'. 

Now,  if  the  bases  of  the  frustum  become  equal,  or 
7?  —  R ',  the  frustum  becomes  a  cylinder,  and  each  of  the 
last  two  terms  in  the  above  expression  for  the  volume  of 
the  frustum  of  a  cone  will  be  equal  to  the  first ;  hence, 
the  volume  of  a  cylinder,  of  which  iZ"is  the  altitude,  and 
It  the  radius  of  the  base,  is  measured  by  H  X  *R2. 

Therefore,  the  volume  of  a  cylinder  is  measured  by  the 
area  of  its  base  multiplied  by  its  altitude. 

Cor.  2.  By  a  process  in  all  respects  similar  to  that  pur- 
sued in  the  case  of  cones,  it  may  be  shown  that  similar 
cylinders  are  to  each  other  as  the  cubes  of  the  radii  of  their 
bases,  and  also  as  the  cubes  of  their  altitudes. 

Cor.  3.  A  cylinder  is  equivalent  to  a  prism  having  an 
equivalent  base  and  an  equal  altitude. 


THEOREM    XXIII. 

If  a  plane  be  passed  through  a  sphere,  the  section  will  be  a 
circle. 

Let  0  be  the  center  of  a  sphere 
through  which  a  plane  is  passed, 
making  the  section  AmBn ;  then 
will  this  section  be  a  circle. 

From  0  let  fall  the  perpendic- 
ular Oo  upon  the  secant  plane, 
and  draw  the  radii  OA,  OB,  and 
Om,  to  different  points  in  the 
intersection  of  the  plane  with 
the   surface  of  the  sphere.     Now, 


212  GEOMETRY. 

the  oblique  lines  OA,  OB,  Om,  are  all  equal,  being  ladii 
of  the  sphere;  they  therefore  meet  the  plane  at  equal  dis 
tances  from  the  foot  of  the  perpendicular  Oo, (Cor.,  Th.  4, 
B.VI);  hence  oA,  oB,  om,  etc.,  are  equal:  that  is,  all  the 
points  in  the  intersection  of  the  plane  with  the  surface  of 
the  sphere  are  equally  distant  from  the  point  0.  This 
intersection  is  therefore  the  circumference  of  a  circle  of 
which  o  is  the  center. 

Hence  the  theorem;  if  a  plane  be  passed  thrcugh  a 
sphere,  etc. 

Cor.  1.  Since  AB,  the  diameter  of  the  section,  is  a  chord 
of  the  sphere,  it  is  less  than  the  diameter  of  the  sphere ; 
except  when  the  plane  of  the  section  passes  through  the 
center  of  the  sphere,  and  then  its  diameter  becomes  the 
diameter  of  the  sphere.     Hence, 

1.  All  great  circles  of  a  sphere  are  equal. 

2.  Of  two  small  circles  of  a  sphere,  that  is  the  greater 
whose  plane  is  the  less  distant  from  the  center  of  the  sphere. 

3.  All  the  small  circles  of  a  sphere  whose  planes  are  at  the 
same  distance  from  the  center,  are  equal. 

Cor.  2.  Since  the  planes  of  all  great  circles  of  a  sphere 
pass  through  its  center,  the  intersection  of  two  great 
circles  will  be  both  a  diameter  of  the  sphere  and  a  com- 
mon diameter  of  the  two  circles.  Hence,  two  great  circles 
of  a  sphere  bisect  each  other. 

Cor.  3.  A  great  circle  divides  the  volume  of  a  sphere,  and 
also  its  surface,  equally. 

For,  the  two  parts  into  which  a  sphere  is  divided  by 
any  of  its  great  circles,  on  being  applied  the  one  to  the 
other,  will  exactly  coincide,  otherwise  all  the  points  in 
their  convex  surfaces  would  not  be  equally  distant  from 
the  center. 

Cor.  4.  The  radius  of  the  sphere  which  is  perpendicular 
to  the  plane  of  a  small  circle,  passes  through  the  center  of  th* 
circ  le. 


BOOK   VII.  213 

Cor.  5.  A  plane  passing  through  the  extremity  of  a  radius 
of  a  sphere,  and  perpendicular  to  it,  is  tangent  to  the  sphere. 

For,  if  the  plane  intersect  the  sphere,  the  section  is  a 
circle,  and  all  the  lines  drawn  from  the  center  of  the 
sphere  to  points  in  the  circumference  are  radii  of  the 
sphere,  and  are  therefore  equal  to  the  radius  which  is  per- 
pendicular to  the  plane,  which  is  impossible,  (Cor.  1,  Th. 
3,  B.  VI).  Hence  the  plane  does  not  intersect  the  sphere, 
and  has  no  point  in  its  surface  except  the  extremity  of 
the  perpendicular  radius.  The  plane  is  therefore  tangent 
to  the  sphere  by  Def  22. 

THEOREM    XXIV. 

If  the  line  drawn  through  the  center  and  vertices  of  two 
opposite  angles  of  a  regular  polygon  of  an  even  number  of 
sides,  be  taken  as  an  axis  of  revolution,  the  perimeter  of  either 
semi-polygon  thus  formed  will  generate  a  surface  whose  measure 
is  the  axis  multiplied  by  the  circumference  of  the  inscribed  circle. 

Let  ABODEF  be  a  semi-polygon  cut 
off  from  a  regular  polygon  of  an  even 
number  of  sides  by  drawing  the  line  AF 
through  the  center  0,  and  the  vertices  A 
and  F,  of  two  opposite  angles  of  the  poly- 
gon ;  then  will  the  surface  generated  by 
the  perimeter  of  this  semi-polygon  re- 
volving about  AF  as  an  axis,  be  meas- 
ured by  AF  X  circumference  of  the  in- 
scribed circle. 

From  m,  the  middle  point,  and  the  extremities  B  and 
C  of  the  side  B  0,  draw  mn,  BK,  and  OL,  perpendicular  to 
AF;  join  also  m  and  0,  and  draw  BIT  perpendicular  to 
QL.  The  surface  of  the  frustum  of  the  cone  generated 
by  the  trapezoid  BKLO,  has  for  its  measure  circ.  mn  X 
BO,  (Cor.  3,  Th.  20).  Since  mO  is  perpendicular  to  BO, 
and  mn  to  BIT,  the  two  A's,  BOH  and  mnO,  are  similar, 
and  their  homologous  sides  give  the  proportion 


214  GEOMETRY. 

mn  :  mO  ::  BE  (=  KL)  :  BO 

and  as  circumferences  are  to  each  other  as  their  radii,  we 
have 

circ.  mn  :  circ.  mO  ::  KL  :  BO 

Hence,     circ.  mn  x  BQ  =  circ.  mO  X  KL. 

But  mO  is  the  radius  of  the  circle  inscribed  in  tne 
polygon.  Hence, the  surface  generated  by  BQ  during  the 
revolution  of  the  semi-polygon,  is  measured  by  the  cir- 
cumference of  the  inscribed  circle  multiplied  by  KL,  the 
part  of  the  axis  included  between  the  two  perpendicu- 
lars let  fall  upon  it  from  the  extremities  B  and  0.  The 
surface  generated  by  any  other  side  of  the  semi-polygon 
will  be  measured,  in  like  manner,  by  the  circumference  of 
the  inscribed  circle  multiplied  by  the  corresponding  part 
of  the  axis. 

By  adding  the  measures  of  the  surfaces  generated  by 
the  several  sides  of  the  semi-polygon,  we  get 

Circ.  mO  x  (AK  +  KL  +  LN+  JSTM+  MF) 

for  the  measure  of  the  whole  surface. 

Hence  the  theorem ;  if  the  line  drawn  through  the  cm 
Mr,  etc. 

Cor.  It  is  evident  that  the  surface  generated  by  any 
portion,  as  CD  and  DK,  of  the  perimeter,  is  measured  by 
circ.  mO  x  LM. 

THEOREM    XXV. 

The  surface  of  a  sphere  is  measured  by  the  circumference 
of  one  of  its  great  circles  multiplied  by  its  diameter. 

Let  a  sphere  be  generated  by  the  revolution  of  the 
Bemi-circle,  AKF,  about  its  diameter,  AF;  then  will  the 
surface  of  the  sphere  be  measured  by 
Circ.  AOxAF. 

Inscribe  in  the  semi-circle  any  regular  semi-polygon, 
and  let  it  be  revolved,  with  the  semi-circle,  about  the  axii 


BOOK   VII.  215 

AF;  the  surface  generated  by  its  perim- 
eter will  be  measured  by 

Circ.  mO  x  AF,  (Th.  24), 

and  this  measure  will  hold  true,  how- 
ever great  the  number  of  sides  of  the  in-  Hj 
scribed  semi-polygon.  But  as  the  num- 
ber of  these  sides  is  increased,  the 
radius  mO,  of  the  inscribed  semi-circle, 
increases  and  approaches  equality  with 
the  radius,  AO;  and  when  we  reach  the  limit,  by 
making  the  number  of  sides  indefinitely  great,  the  radii 
and  semi-circles  become  equal,  and  the  surface  generated 
by  the  perimeter  of  the  inscribed  semi-polygon  becomes 
the  surface  of  the  sphere.  Therefore,  the  surface  of  the 
sphere  has,  for  its  measure, 

Circ.  A  0  x  AF. 

Hence  the  theorem ;  the  surface  of  a  sphere  is  meas- 
ured, etc. 

Cor.  1.  A  zone  of  a  sphere  is  measured  by  the  circumfer- 
ence of  a  great  circle  of  the  sphere  multiplied  by  the  altitude 
of  the  zone. 

For,  the  surface  generated  by  any  portion,  as  CD  and 
DE,  of  the  perimeter  of  the  inscribed  semi-polygon  has, 
for  its  measure,  circ.  mO  X  LM,  (Cor.  Th.  24) ;  and  as 
the  number  of  the  sides  of  the  semi-polygon  increases, 
LM  remains  the  same,  the  radius  mO  alone  changing, 
and  becoming,  when  we  reach  the  limit,  equal  to  AO: 
hence,  the  surface  of  the  zone  is  expressed  by 

Circ.  Ad  X  LM, 
whether  the  zone  have  two  bases,  or  but  one. 

Cor.  2.  Let  H  and  Hr  denote  the  altitudes  of  two 
zones  of  spheres,  whose  radii  are  R  and  R ' ;  then  these 
zones  will  be  expressed  by  2<>rR  x  H  and  2*R '  x  Rr; 
and  if  tht  surfaces  of  the  zones  be  denoted  by  Z  and  Z'f 
we  have 


216  GEOMETRY. 

Z  :  Z1  :    2«RxH  :  2«R'  x  Hf  i:  Rx  E  :  R'  x  H'> 

•Hence,  1.  Zones  in  different  spheres  are  to  each  other  at 
their  altitudes  multiplied  by  the  radii  of  the  spheres. 

2.  Zones  of  equal  altitudes  are  to  each  other  as  the  radii 
of  the  spheres, 

3.  Zones  in  the  same,  or  equal  spheres,  are  to  each  other  as 
their  altitudes. 

Cor.  3.  Let  R  denote  the  radius  of  a  sphere;  then  will 
its  diameter  be  expressed  by  222,  and  the  circumference 
of  a  great  circle  by  2irR;  hence  its  surface  will  be  ex 
pressed  by 

2«R  x  2R  =  4iri22. 

That  is,  the  surface  of  a  sphere  is  equivalent  to  the  area  of 
four  of  its  great  circles. 

Cor.  4.  The  surfaces  of  spheres  are  to  each  other  as  the 
squares  of  their  radii. 

THEOREM    XXVI. 

Tf  a  triangle  be  revolved  about  either  of  its  sides  as  an  axis, 
the  volume  generated  will  be  measured  by  one  third  of  the  prod- 
uct of  the  axis  and  the  area  of  a  circle,  having  for  its  radius 
the  perpendicular  let  fall  from  the  vertex  of  the  opposite 
angle  on  the  axis,  or  on  the  axis  produced. 

First.  Let  the  triangle  ABC, 
in  which  the  perpendicular  from 
C  falls  on  the  opposite  side,  AB, 
be  revolved  about  AB  as  an  axis ; 
then  will  *Vol.  A  ABChuve,  for 
its  measure,  \AB  x  *CD . 

The  two  A's  into  which  A  ABC  is  divided  by  the 
perpendicular  DC,  are  right-angled,  and  during  the  rev- 
olution they  will  generate  two  cones,  having  for  their 

*  Vol.  A  ABC,  cone  A  ADC,  are  abbreviations  for  volume  gener- 
ated by  A  ABC,  cone  generated  by  A  ADC;  and  surfaces  of  revolu- 
sien  generated  by  lines  will  hereafter  be  denoted  by  like  abbreviations. 


BOOK   VII.  217 

common  base  the  circle,  of  which  D  0  is  the  radius,  and 
for  their  axes  the  parts  DA  and  DB,  into  which  AB  is 
divided. 

Now,  *Cone  a  ADO  is  measured  by  \AD  x  *DO\ 
(Th.  21),  and  cone  A  BDO,  by  %BD  x  «DO%  \  but  these 
two  cones  compose  Yol.  A  ABO;  and  by  adding  their 
measures,  we  have,  for  that  of  Yol.  A  ABO, 

IAD  x  *D02  +  iBD  x  *~D(72  =  J  AB  x  <^DO\ 
Second.  Let  the  trian-  9 

gle  FFGr,  in  which  the 
perpendicular  from  G- 
falls  on  the  opposite  side 
FF  produced,  be  revolved 
about  FF   as    an    axis ; 

then  will  Yol.  A  FFG    E    F_         "h 

have,  for  its  measure,  %EF  x  *GfH\  CrH  being  the  per- 
pendicular on  FF  produced.  For,  in  this  case  it  is  appa- 
rent, that  Yol.  A  FFGr  is  the  difference  between  the 
cone  A  FHCr  and  the  cone  a  FHCr.  The  first  cone  has, 
for  its  measure,  \FH  x  *GH\  and  the  second,  for  its 
measure,  %FH  x  wGrJI2 ;  hence,  by  subtraction,  we  have 

Vol.  A  FFG  =  IEH X  hGH2  —  iFH X  7t~GH2  =  IEF X  7t~GH2. 
Hence  the  theorem ;  if  a  triangle  be  revolved  about  either 
of  its  sides,  etc. 

Scholium. — If  we  take  either  of  the  above  expressions  for  the  meas- 
ure of  the  volume  generated  by  the  revolution  of  a  triangle  about  one 
of  its  sides,  for  example  the  last,  and  factor  it  otherwise,  we  have 

\EF  X  *GH*  =  EFX  $GHxU*2GH=  FFx  $GHX  — 3 

Now,  EF  X  %GH  expresses  the  area  of  the  triangle  EFG;  and 

2x  X  GH 

,  one  third  of  the  circumference  described  by  the  point  Q 

o 

during  the  revolution. 

The  expression,  \AB  X  rtDC2,  maybe  factored  and  interpreted  in  the 

*  See  note  on  the  preceding  page. 
19 


218 


GEOMETRY. 


same  manner.  Hence,  we  conclude  that  the  volume  generated  by  th* 
revolution  of  a  triangle  about  either  of  its  sides,  is  measured  by  the  area 
of  the  triangle  multiplied  by  one  third  of  the  circumference  described  in 
the  revolution  by  the  vertex  of  the  angle  opposite  the  axis. 


THEOREM    XXVII. 

The  volume  generated  by  the  revolution  of  a  triangle  about 
any  line  lying  in  its  plane,  and  passing  through  the  vertex  of 
one  of  its  angles,  is  measured  by  the  area  of  the  triangle  mul- 
tiplied by  two  thirds  of  the  circumference  described,  in  the 
revolution,  by  the  middle  'point  of  the  side  opposite  the  vertex 
through  which  the  axis  passes. 

Let  the  triangle  ABCbe 
revolved  about  the  line 
AG,  drawn  through  the 
vertex  A,  and  lying  in  the 
plane  of  the  triangle,  and 
let  HE  be  the  perpendicu- 
lar let  fall  from  H,  the 
middle  point  of  BO,  upon 
the  axis  AG- ;  then  will  Vol.  a  ABO  have,  for  its  meas 
ure,  A  ABO  x  §  circ.  HE. 

From  the  extremities  of  BO,  let  fall  the  perpendicu- 
lars BE  and  OB,  on  the  axis;  and  from  A  draw  AST  per 
pendicular  to  BO,  or  BO  produced,  and  produce  OB, 
until  it  meets  the  axis  in  G-. 

Now,  it  is  evident  that  Yol.  A  ABO  is  the  difference 
between  Yol.  A  AGO  and  Yol.  A  AGB.  But  Yol. 
A  AGO  is  expressed  by  a  AGO  x  J  circ.  (7i>;and  Yol. 
A  AGB,  by  A  AG B  x  J  circ.  BE,  (Scholium,  Th.  26). 
Hence, 

Vol.  A  ABC  =  ^  AGO  X  I  circ.  CD  —  A  AGB  X  i  circ.  BF. 

Substituting  for  areas  of  A's,  and  for  circumferences, 
their  measures,  we  have 


BOOK   VII.  219 

Vol.  A  A£C=  GO  X  IAK  X  ?^2—  GB  x  UK  X  ^^ 

o  o 

=  GCx  \AKx  2J^R—{GC—BC)  X  UKX  ^^ 

o  • 

~GCXIAKX^^—  GCXIARX^^+BCXUK*^1^ 
o  8  8 

=  GO  X  \AK  X  ^(CD  —  BF)  +  BC  X  $AR  X  — „— . 

o  o 

But  i?iV  being  drawn  parallel  to  J.#,  we  have 
GN  =  CD  —  BF; 
hence,  substituting  this  value  for  CD  —  BF,  in  the  first 
term  of  the  second  member  of  the  last  equation,  we  have 

o  o 

=  GCx  CNx  ±AK  x2^  +  BC  x  %AK  x  ?^^, 

by  changing  the  order  of  factors  in  the  first  term  of  the 
second  member.  The  homologous  sides  of  the  similar 
triangles,  GOD  and  BCN,  give  the  proportion 

GO  :  CD  : :  BO  :  ON 

whence,  GO  x  ON  =  CD  x  BO 

Substituting  this  value  for  GO  X  CJSF,  in  the  last  equa- 
tion above,  and  arranging  the  factors  as  before,  it  becomes 

Vol.  A  ABC=  BO  x  lAKx  ^^  +  BC x  \AK x  ^F 
-  BO  x  UK  x  g^V  BF>. 

a 

But  CD  +  BF=  2HF;  hence 
Vol,  A  ABC=--BCx  \AKx  —~=BCx  ±AKx  f  .2*.  J7#j 

o 

and  since 

BO  x  \AK=  A  ABC,  and  §  x  2«.HE  -  §  circ.  ##, 

this  measure  conforms  to  the  enunciation. 

It  only  remains  for  us  to  consider  the  case  in  which 
the  axis  is  parallel  to  the  base  BC  of  the  triangle      The 


220 


GEOMETKY. 


precedi  :g  demonstration  will  not  now  apply,  because  it 
supposes  BO,  or  B 0  produced,  to  intersect  the  axis. 

Let  the  axis  AE,  be  parallel  to  the 
base  BO,  of  the  A  ABO.  From  B 
and  0  let  fall  on  the  axis  the  perpen- 
diculars BE  and  OB. 

Now  it  is  plain  that 

Vol.  A  ABO=  cylinder  rectangle  BODE  -f 
cone  a  ADO —  cone  A  AEB. 

Substituting  in  second  member,  for  cylinder  and  cones, 
their  measures,  we  have 

Vol.  AABO=  BE  x  *£Z)2  +  IAD  x  ^OD2—  \AE  x  «BE* 
=$DEx  *7JD2+iDEx  «OD2+§ADx*OD*—iAEx  *BE\ 

But  BE  =  OD,  and  ^DE  +  %AD  =  \AE.  Reducing  by 
these  relations,  we  have 

Vol.  A  ABO=  %DE  x  «OD3=  %DE  x  \OD  x  4*.OD 
=  DEx  lODx  %.<L«.0D  =  B0x  \QD  x  \Z«.OD. 

And,  since  BO  x  \OD  expresses  the  area  of  the  tri- 
angle ABO,  and  §.2^.(77),  two  thirds  of  the  circumfer- 
ence described  by  any  point  of  the  base,  this  expression 
also  conforms  to  the  enunciation. 

Hence  the  theorem ;  the  volume  generated  by  the  revela- 
tion, etc. 

Oor.  If  the  generating 
triangle  becomes  isosceles, 
the  perpendicular  from  A 
meets  the  base  at  its  middle 
point.  In  this  case,  if  we 
resume  the  expression 

BOx  \AK  x4-^. 

it  becomes 

BO  x  \AK  x  KE  x  i*. 


BOOK    VII.  221 

But,  since  AKis  perpendicular  to  BC,  and  KE  to  BN, 
the  a's  AKE  and  CBN  are  similar,  and  their  honiolo* 
gous  sides  give  the  proportion 

BO  :  BJST  ::  AK  :  KE 

whence,  BCx  KE  =  BJSTx  AK 

Changing  the  order  of  factors  in  the  last  expression  on 
the  preceding  page,  and  replacing  BOxKE  by  its  value, 
it  becomes 

lAKx  AKxBNx  ±*  =  AK2  xMxf* 

Hence, 

Vol.  A  ABO=  |*  x  AK7  x  BN.=  $«  x  AK2  x DF 

That  is,  the  volume  generated  by  the  revolution  of  an  isos  - 
celes  triangle  about  any  line  drawn  through  its  vertex  and  lying 
in  the  plane  of  the  triangle,  is  measured  by  \n  times  the  square 
of  the  perpendicular  of  the  triangle  multiplied  by  the  part  of  the 
axis  included  between  the  two  perpendiculars  let  fall  upon  it 
from  the  extremities  of  the  base  of  the  triangle. 

Scholium. — If  we  resume  the  equation 

Vol.  A  ABC  =  BC  X  \AK  X  ^^ 

o 

and  change  the  order  of  the  factors  in  the  second  member,  it  may  be 
put  under  the  form 

Vol.  A  ^BC  =  BC  X  2*.HE  X  \AK. 

But  during  the  revolution  of  the  triangle,  the  side  BC  generates  the 
surface  of  the  frustum  of  a  cone,  which  surface  has  for  its  measure 

BC  X  2*.HE  (Th.  20,  Cor.  3). 

Hence,  the  above  equation  may  be  thus  interpreted:  The  volumt 
generated  by  the  revolution  of  a  triangle  about  any  line  lying  in  its  plans 
and  passing  through  the  vertex  of  one  of  its  angles,  is  measured  by  the 
surface  generated,  during  the  revolution,  by  the  side  opposite  the  vertex 
through  which  the  axis  passes  multiplied  by  one  third  of  the  perpen- 
dicular drawn  from  the  vertex  to  that  side. 


19* 


222 


GEOMETRY. 


THEOREM    XXVIII. 

If  the  line  drawn  through  the  center  and  vertices  of  two  op* 
posite  angles  of  a  regular  polygon,  of  an  even  number  of 
sides,  be  taken  as  an  axis  of  revolution,  either  semi-polygon 
thus  formed  will,  during  this  revolution,  generate  a  volume 
which  has,  for  its  measure,  the  surface  generated  by  the 
perimeter  of  the  semi-polygon  multiplied  by  one  third  of  its 
apothem. 

Let  ABODE  be  a  regular  semi-poly- 
gon, cut  off  from  a  regular  polygon 
of  an  even  number  of  sides,  by  draw- 
ing a  line  through  the  center,  0,  and 
the  vertices,  A  and  E,  of  two  opposite 
angles  of  the  polygon ;  then  will  the 
volume  generated  by  the  revolution 
of  this  semi-polygon  about  AE,  as  an 
axis,  be  measured  by  (Sur.  AB  -f  sur. 
BO  +  sur.  CD  -f  sur.  DE)  x  }Om,  Om 
being  the  apothem  of  the  polygon. 

For,  if  from  the  center  of  0,  the  lines  OB,  00,  OD,  be 
drawn  to  the  vertices  of  the  several  angles  of  the  semi- 
polygon,  it  will  be  divided  into  equal  isosceles  triangles, 
the  perpendicular  of  each  being  the  apothem  of  the 
polygon. 

Now,  the  volume  generated  by  A  AOB  has,  for  its 
measure, 

Sur.  AB  x  iOm, 
Sur.  BO  X  iOm, 
Sur.  OD  x  \Om, 


that  by  A  BOO, 
"  A  OOD, 
"      A  DOE, 


Sur.  DE  x  iOm,  (Scholium,  Th.  27). 


By  the  addition  of  the  measures  of  these  partial  vob 
nines,  we  find,  for  that  of  the  whole  volume, 

Vol.  semi-polygon  ABODE  =  sur.  perimeter  ABODE  X  iOm, 

and  were  the  number  of  the  sides  of  the  semi -polygon 


BOOK  VII.  223 

increased  or  diminished,  the  reasoning  would  be  in  no 
wise  changed. 

Hence  the  theorem ;  if  the  line  drawn  through  the  cen- 
ter, etc. 

Scholium. — The  volume  generated  by  any  portion  of  the  semi-poly- 
gon, as  that  composed  of  the  two  isosceles  /±'a  BOC,  COD,  is  meas 

ured  by 

Sur.  perimeter  BCD  X  10m. 

THEOREM    XXIX. 

The  volume  of  a  sphere  is  measured  by  its  surface  multi- 
flied  by  one  third  of  its  radius. 

Let  a  sphere  be  generated  by  the 
revolution  of  the  semicircle  AOE, 
about  its  diameter,  AE,  as  an  axis; 
then  will  the  volume  of  the  sphere  be 
measured  by 

sur.  semi-circ.  OA  x  \OA. 

For,  inscribe  in  the  semi-circle  any 
regular  semi  -  polygon,  as  ABODE, 
and  let  it,  together  with  the  semi-cir- 
cle, revolve  about  the  axis  AE.  The 
semi-polygon  will  generate  a  volume  which  has,  for  i  ts 
measure, 

Sur.  perimeter  ABODE  x  JOw,  (Th.  28), 
in  which  Om  is  the  apothem  of  the  polygon. 

Now,  however  great  the  number  of  sides  of  the  in- 
scribed regular  semi-polygon,  this  measure  for  the  volume 
generated  by  it,  will  hold  true ;  but  when  we  reach  the 
limit,  by  making  the  number  of  sides  indefinitely  great, 
the  perimeter  and  apothem  become,  respectively,  the 
semi-circumference  and  its  radius,  and  the  volume  gen 
erated  by  the  semi-polygon  becomes  that  generated  by 
the  semi-circle,  that  is,  the  sphere.     Therefore, 

Vol.  sphere  =  sur.  semi-circ.  OA  x  %OA. 


224 


GEOMETKY. 


Scholium  1. — If  we  take  any  portion  of  the  inscribed  senii-pulygon, 
as  BO C,  the  volume  generated  by  it  is  measured  by  sur.  BC  X  %Omy 
(Scholium,  Th.  27) ;  and  when  we  pass  to  the  limit,  this  volume  be- 
comes a  sector,  and  sur.  BC  a  zone  of  the  sphere,  which  zone  is  the 
base  of  the  sector.  Hence,  the  volume  of  a  spherical  sector  is  measured 
by  the  zone  which  forms  its  base  multiplied  by  one  third  of  the  radiux 
of  the  sphere. 

Scholium  2.  —  Let  R  denote  the  radius  of  a  sphere ;  then  will  its 
diameter  be  represented  by  2R.  Now,  since  the  surface  of  a  sphere  is 
equivalent  to  the  area  of  four  of  its  great  circles,  and  the  area  of  a 
great  circle  is  expressed  by  7tR2,  we  have 

Vol.  sphere  =  4*R*  X  IR  =  \rtRK 

And  since  R3  =  l(2R )3,  we  also  have 

Vol.  sphere  =  £*i23  =  i*(2i?)». 

Hence,  the  volume  of  a  sphere  is  measured  by  four  thirds  of  it  times  the 
cube  of  the  radius,  or  by  one  sixth  of  tt  times  the  cube  of  the  diameter. 

THEOREM    XXX. 

The  surface  of  a  sphere  is  equivalent  to  two  thirds  of  the 
surface,  bases  included,  and  the  volume  of  a  sphere  to  two 
thirds  of  the  volume,  of  the  circumscribing  cylinder. 

Let  AMD  be  a  semi-circle,  and 
ABQD  a  rectangle    formed  by  B: 
drawing  tangents    through    the 
middle  point  and  extremities  of 
the  semi-circumference,  and  let  Mr- 
the  semi-circle  and  rectangle  be 
revolved  together  about  AD  as 
an  axis.     The  rectangle  will  thus  c 
generate  a  cylinder  circumscribed 
about  the  sphere  generated  by  the  semi-circle. 

First.  The  diameter  of  the  base,  and  the  altitude  of 
the  cylinder,  are  each  equal  to  the  diameter  of  the 
sphere ;  hence  the  convex  surface  of  the  cylinder,  being 
measured  by  the  circumference  of  its  base  multiplied  by 
its  altitude,  (Cor.  1,  Th.  20),  has  the  same  measure  aa 
the  surface  of  the  sphere,  (Th.  25).  But  the  surface  of 
the  sphere  is  equivalent  to  four  great  circles,  (Cor.  3j 


/^^        A             "^\ 

T^z 

H 

^- 

-%i 

J?N 

^^^_    D     _^s 

BOOR   VII.  225 

Th.  25).  Hence,  the  convex  surface  of  the  cylinder  is 
equivalent  to  four  great  circles ;  and  adding  to  these  the 
bases  of  the  cylinder,  also  great  circles,  we  have  the 
whole  surface  of  the  cylinder  equivalent  to  six  great 
circles.  Therefore,  the  surface  of  the  sphere  is  four 
sixths  =  two  thirds  of  the  surface  of  the  cylinder,  in- 
cluding its  bases. 

Second.  The  volume  of  the  cylinder,  being  measured 
by  the  area  of  the  base  multiplied  by  the  altitude,  (Cor. 
1,  Th.  22),  is,  in  this  case,  measured  by  the  area  of  a 
great  circle  multiplied  by  its  diameter  =  four  great  cir- 
eles  multiplied  by  one  half  the  radius  of  the  sphere. 

But  the  volume  of  the  sphere  is  measured  by  four 
great  circles  multiplied  by  one  third  of  the  radius,  (Scho- 
lium 2,  Th.  29).    Therefore, 

Vol.  sphere  :  Vol.  cylinder  : :  J  :  J  : :  2  :  3 ; 

whence,     Vol.  sphere  =  §  Vol.  cylinder. 

Hence  the  theorem ;  the  surface  of  a  sphere  is  equiva- 
lent, etc. 

Cor.  The  volume  of  a  sphere  is  to  the  volume  of  the  cir~ 
cumscribed  cylinder,  as  the  surface  of  the  sphere  is  to  the  sur- 
face of  the  cylinder. 

Scholium. — Any  polyedron  circumscribing  a  sphere,  may  be  regarded 
as  composed  of  as  many  pyramids  as  the  polyedron  has  faces,  the  cen- 
ter of  the  sphere  being  the  common  vertex  of  these  pyramids,  and  the 
several  faces  of  the  polyedron  their  bases.  The  altitude  of  each  pyra- 
mid will  be  a  radius  of  the  sphere  ;  hence  the  volume  of  any  one  pyra- 
mid will  be  measured  by  the  area  of  the  face  of  the  polyedron  which 
forms  its  base,  multiplied  by  one  third  of  the  radius  of  the  sphere. 
Therefore,  the  aggregate  of  these  pyramids,  or  the  whole  polyedron, 
will  be  measured  by  the  surface  of  the  polyedron  multiplied  by  one 
third  of  the  radius  of  the  sphere. 

But  the  volume  of  the  sphere  is  also  measured  by  the  surface  of  the 
sphere  multiplied  by  one  third  of  its  radius.     Hence, 

Sur.  polyedron  :  Sur.  sphere  : :  Vol.  polyedron  :  Yol.  sphere. 

That  is,  the  surface  of  any  circumscribed  yolycdron  is  to  the  surface 
of  the  sphere^  as  the  volume  of  the  polyedrori  is  to  the  volume  of  ihs 
sphere. 


22t$  GEOMETR  Y. 

THEOREM    XXXI. 

The  volume  generated  by  the  revolution  of  the  segment  of  a 
circle  about  a  diameter  of  the  circle  exterior  to  the  segment,  is 
measured  by  one  sixth  of  «r  times  the  square  of  the  chord  of 
the  segment,  multiplied  by  the  part  of  the  axis  included  be- 
tween the  perpendiculars  let  fall  upon  it  from  the  extremities 
of  the  chord, . 

Let  BOB  be  a  segment  of  the  circle, 
whose  center  is  0,  and  AH  a  part  of  a 
diameter  exterior  to  the  segment.  Draw 
the  chord  BB,  and  from  its  extremities 
let  fall  the  perpendiculars,  BF,  BE  on 
AH;  also  draw  Om  perpendicular  to 
BB.  The  spherical  sector  generated 
by  the  revolution  of  the  circular  sector 
BCBO  about  AH,  is  measured  by  zone  BB  x  \B0, 
(Scholium  1,  Th.  29),  =  2«.BO  x  EF  X  %BO  -  §*W  x 
EF;  and  the  volume  generated  by  the  isosceles  triangle 
BOB  is  measured  by 

\^Om"  x  EF,  (Cor.  1,  Th.  27). 

The  difference  between  these  two  volumes  is  that  gen- 
erated by  the  circular  segment  BOB,  which  has,  there- 
fore, for  its  measure, 

l«EF(BO*  —  ~Om)  -  %«EF  x  Bm^,  (Th.  39,  B.  I). 

But  since  Bm  =  \BB,  Bm   =  \BB*  \  hence,  by  sub- 
Btituting,  we  have 
Vol.  segment  BOB  -  \«EF  x  \BB*  -  \*~BB*  x  EF. 

Hence  the  theorem. 

THEOREM    XXXII. 

The  volume  of  a  segment  of  a  sphere  has,  for  its  measure, 
the  half  sum  of  the  bases  of  the  segment  multiplied  by  its  alti- 
tude, plus  the  volume  of  a  sphere  which  has  this  altitude  for 
its  diameter. 


BOOK   V  II.  227 

Let  BOB  be  the  arc  of  a  circle,  and 
BF  and  BE  perpendiculars  let  fall 
from  its  extremities  upon  a  diameter,  cA 
of  which  AH  is  a  part;  then,  if  the 
area  BCBEF  be  revolved  about  AH  I 
as  an  axis,  a  spherical  segment  will 
be  generated,  for  the  volume  of  which 
it  is  proposed  to  find  a  measure. 

The  circular  segment  will  generate  a  volume  meas- 
ured by  \«BB2  x  EF,  (Th.  31) ;  and  the  frustum  of  the 
cone  generated  by  the  trapezoid  BBEF  will  have,  for 
its  measure, 

i<r"5P2  x  EF+  i«BE*  xEF+  i«BF  xBEx  EF,  (Th.  22), 
=  §«EF(BF2  +~BE2  +  BF  x  BE). 

But  the  sum  of  these  two  volumes  is  the  volume  of 
the  spherical  segment,  which  has,  therefore,  for  its 
measure, 

i*EF  {BB2  +  2BF2  +  2BE2  +  2BF  x  BE) 
From  B  let  fall  the  perpendicular  Bn  on  BE;  then  will 
Bn  =  BE—nE=BE  —  BF; 


hence,      Bn  =  BE2  —  2BE  x  BF  +  BF2 


2  "^—2 


and  since    BB   -  Bn   +  Bn    =  EF  +  Bn  % 

we  have    BB2  =  EF2 +  BE2 +  BF2 —  2BE  x  BF. 

By  substituting  this  value  for  BB2,  in  the  above  meas- 
ure for  the  volume  of  the  segment,  we  find 

UEF(EF2  +  DE2  +  RF2^2DExBF^2BF2-j-2DE2  +  2BFxDE] 
r-  I  «EF  {EF2  +  WE2  +  ZBF2)  =faEF*  +  EF  (^DE  ~^—--  ). 

Which  last  expression  conforms  to  the  enunciation. 

Hence  the  theorem ;  the  volume  of  a  segment  of  a  sphere, 
etc. 

Cor.  When  the  segment  has  but  one  base,  BF  becomes 
eero,  and  EF  becomes  EA\   and  the  final   expression 


228  GEOMETRY. 

which  we  found  for  the  volume  of  the  segment  reduces 
to 

$«£A>  +  EA  x  ' 


2 

Hence,  A  spherical  segment  having  but  one  base,  is  equiva- 
lent to  a  sphere  whose  diameter  is  the  altitude  of  the  segment, 
plus  one  half  of  a  cylinder  having  for  base  and  altitude  the 
base  and  altitude  of  the  segment. 

Scholium. — "When  the  spherical  segment  has  a  single  base,  we  may 

put  the  expression,  \hEA   +  EA  X  — — ,  under  a  form  to  indicate  a 

convenient  practical  rule  for  computing  the  volume  of  the  segment. 

Thus,  since  the  triangle  DEO  is  right-angled,  and  0j£=  OA  —  EA, 
we  have 

DE2  =TDOa  — ~OE2  =  O?  —  Ol2  +  20A  X  EA—~EA2 

=  20Ax  EA—~EA\ 

By  substituting  this  value  for  DE2  in  the  expression  for  the  volume 
of  the  segment,  we  find 

UEA?  +  EAX?X(20AXEA  —  ~EA*) 

2 

Bsi*1Zf  -f  EA2  X  %  [20A  —  EA) 

2 

=  }rtEA9+U.ZEAi  {20A  —  EA) 
=  \hEA\EA  +  Q.OA  —  3  EA) 
=  lrtEA\6.0A  —  2EA) 
=  l7tEA\%OA  —  EA) 

Hence,  the  volume  of  a  spherical  segment,  having  a  single  base,  is 
measured  by  one  third  of  n  times  the  square  of  the  altitude  of  the  seg- 
ment, multiplied  by  the  difference  between  three  times  the  radius  of  lfa& 
sphere  and  this  altitude. 

RECAPITULATION 

Of  some  of  the  principles  demonstrated  in  this  and  the  pre- 
ceding Boohs. 

Let  R  denote  the  radius,  and  D  the  diameter  of  any 
circle  or  sphere,  and  H  the  altitude  of  a  cone,  or  of  a 
segment  of  a  sphere ;  then, 


-i*r +  *&*"  + **"•> 


BOOK  VII.  229 

Circuinforence  of  a  circle         ==  2*JK. 
Surface  of  a  sphere  =  4*J22,  or  nD\ 

Zone  forming  the  base  of  a  1    _  ~  ^       „. 

segment  of  a  sphere,         / 
Volume  or  solidity  of  a  sphere  =  {*B*,  or  £*!)•. 
Volume  of  a  spherical  sector  =  %nR2  x  JZ". 
Volume  of  a  cone,  of  which  ^ 

Jit  is  the  radius  of  the  V  =  \*IP  x  H. 

base  ) 

Volume  of  a  spherical  seg-' 

ment,  of  which  R'  is  the 

radius  of  one  base,  and 

R"    the   radius    of   the 

other,  and  whose  altitude 

is#, 

If  the  so ir ment  has  but  one^        .    „,  ,    ^hR'* 

;,„  -,  ,!        =  \*H*  +  H.—^  ;  or, 

base,  R"  =  zero,  and  the  >  2 

volume  of  the  segment,   J   =  J*JP(3i2  —  H). 

PRACTICAL    PROBLEMS. 

1.  The  diameter  of  a  sphere  is  12  inches  ;  how  many 
cubic  inches  does  it  contain?  Arts.  904.78  cu.  in. 

2.  What  is  the  solidity  of  the  segment  of  a  single  base 
that  is  cut  from  a  sphere  12  inches  in  diameter,  the  altitude 
of  the  segment  being  3  inches?       Arts.  141.372  cu.  in. 

3.  The  surface  of  a  sphere  is  68  square  feet ;  what  is 
its  diameter  ?  Ans.  D  =  4.652  feet. 

4.  If  from  a  sphere,  whose  surface  is  68  square  feet,  a 
segment  be  cut,  having  a  depth  of  two  feet  and  a  single 
base,  what  is  the  convex  surface  of  the  segment  ? 

Ans.  29.229+  sq.  ft. 

5.  What  is  the  solidity  of  the  sphere  mentioned  in  the 
two  {receding  examples,  and  what  is  the  solidity  of  the 
segment,  having  a  depth  of  two  feet,  and  but  one  base  ? 

A        (  Solidity  of  sphere,     52.71  cu.  ft. 
t        "        "   segment,  20.85       " 
20 


230  GEOMETRY. 

6.  In  a  sphere  whose  diameter  is  20  feet,  what  is  the 
solidity  of  a  segment,  the  bases  of  which  are  on  the  same 
side  of  the  center,  the  first  at  the  distance  of  3  feet  from 
it,  and  the  second  of  5  feet;  and  what  is  the  solidity  of 
a  second  segment  of  the  same  sphere,  whose  bases  are 
also  on  the  same  side  of  the  center,  and  at  distances 
from  it,  the  first  of  5  and  the  second  of  7  feet  ? 

a       (  Solidity  of  first  segment,  525.7  cu.  ft 
US  I        "        "  second    "        400.03     " 

7.  If  the  diameter  of  the  single  base  of  a  spherical 
segment  be  16  inches,  and  the  altitude  of  the  segment  4 
inches,  what  is  its  solidity  ?  * 

Arts.  435.6352  cubic  inches. 

8.  The  diameter  of  one  base  of  a  spherical  segment  is 
18  inches,  and  that  of  the  other  base  14  inches,  these 
bases  being  on  opposite  sides  of  the  center  of  the  sphere, 
and  the  distance  between  them  9  inches ;  what  is  the 
volume  of  the  segment,  and  the  radius  of  the  sphere  ? 

a         (  Vol.  seg.,  2219.5  cubic  inches. 
\  Bad.  of  sphere,  9.4027  inches. 

9.  The  radius  of  a  sphere  is  20,  the  distance  from  the 
center  to  the  greater  base  of  a  segment  is  10,  and  the 
distance  from  the  same  point  to  the  lesser  base  is  16 ; 
what  is  the  volume  of  the  segment,  the  bases  being  on 
the  same  side  of  the  center?  Ans.  4297.7088. 

10.  If  the  diameter  of  one  base  of  a  spherical  segment 
be  20  miles,  and  the  diameter  of  the  other  base  12  miles* 
and  the  altitude  of  the  segment  2  miles,  what  is  its 
solidity,  and  what  is  the  diameter  of  the  sphere  ? 

*  First  find  the  radius  of  the  sphere. 

Note. — The  Key  to  this  work  contains  full  solutions  to  all  the  problems  in 
the  Geometry  and  Trigonometry,  and  the  necessary  diagrams  for  illustration. 


BOOK  VIII.  231 


BOOK  VIII. 


PRACTICAL  GEOMETRY. 

APPLICATION    OF    ALGEBRA   TO   GEOMETRY,   AND    ALSO 
PROPOSITIONS  FOR  ORIGINAL  INVESTIGATION. 

No  definite  rules  can  be  given  for  the  algebraic  solu- 
tion of  geometrical  problems.  The  student  must,  in  a 
a  great  measure,  depend  on  his  own  natural  tact,  and 
Lis  power  of  making  a  skillful  application  of  the  geomet- 
rical and  analytical  knowledge  he  has  thus  far  obtained. 

The  known  quantities  of  the  problem  should  be  repre- 
sented by  the  first  letters  of  the  alphabet,  and  the  un- 
known by  the  final  letters ;  and  the  relations  between 
these  quantities  must  be  expressed  by  as  many  inde- 
pendent equations  as  there  are  unknown  quantities.  To 
obtain  the  equations  of  the  problem,  we  draw  a  figure, 
the  parts  of  which  represent  the  known  and  unknown 
magnitudes,  and  very  frequently  it  will  be  found  neces- 
sary to  draw  auxiliary  lines,  by  means  of  which  we  can 
deduce,  from  the  conditions  enunciated,  others  that  can 
be  more  conveniently  expressed  by  equations.  In  many 
cases  the  principal  difficulty  consists  in  finding,  from  the 
relations  directly  given  in  the  statement,  those  which 
are  ultimately  expressed  by  the  equations  of  the  problem. 
Having  found  these  equations,  they  are  treated  by  the 
known  rules  of  algebra,  and  the  values  of  the  required 
magnitudes  determined  in  terms  of  those  given. 


23:2 


GEOMETRY. 


PROBLEM    I. 

Given,  the  hypotenuse,  and  the  sum  of  the  other  two  sides 
of  a  right-angled  triangle,  to  determine  the  triangle. 

Let  ABO  be  the  A.  Put  OB  =  y,  AB 
=  x,  AO=  h,  and  OB  +  AB  =  s.  Then, 
by  a  given  condition,  we  have 

x  +  y  =  s; 
and,  x*+  y*=  h\  (Th.  39,  B.  I). 

lleducing  these  two  equations,  and  we  have 

x  =  \s  =b  i^W^7;         y  mm  \s  =fc  Jv^tf  —  *2. 

If  A  =  5  and  *  =  7,  a;  =  4  or  3,  and  y  =  3  or  4. 

Remark.  —  In  place  of  putting  x  to  represent  one  side,  and  y  the 
other,  we  might  put  [x  -f-  y)  to  represent  the  greater  side,  and  (x  —  y) 
the  less  side ;  then, 

h2 
x*  +  y*  =  -,  and  2x  =  s,  etc. 


PROBLEM    II. 

(riven,  the  base  and  perpendicular  of  a  triangle,  to  find  the 
side  of  its  inscribed  square. 

Let  ABO  be  the  A.  Put 
AB  mm  b,  the  base,  OB  —  p, 
the  perpendicular. 

Draw  EF  parallel  to  AB, 
and  suppose  it  equal  to  EG, 
a  side  of  the  required  square ;  and  put  EF  =  x. 

Then,  by  similar  A's,  we  have 

01 :  EF  : :  OB  :  AB. 


That  is, 
Hence, 


p  —  x 


P 


b. 


bp  —  bx  =  px ;  or,  x  m  t    f   . 
r         -  b  +  p 


That  is,  the  side  of  the  inscribed  square  is  equal  to  the 
product  of  the  base  and  altitude,  divided  by  their  sum. 


BOOK   VIII 


233 


PROBLEM    III. 

In  a  triangle,  having  given  the  sides  about  the  vertical 
angle,  and  the  line  hisecting  that  angle  and  terminating  in 
the  base,  to  find  the  base. 

Let  ABO  be  the  a,  and  let  a  cir- 
cle be  circumscribed  about  it.  Di- 
vide the  arc  AEB  into  two  equal 
partu  at  the  point  E,  and  draw  EO. 
This  line  bisects  the  vertical  angle, 
(Cor.,  Th.  9,  B.  HI).     Draw  BE. 

Put  AD  =  x,  DB  =  y,  AQ  =  a, 
OB  =  b,  OD  =  c,  and  BE  =  w.     The  two  A's,  ADO  and 
EBO,  are  equiangular;  from  which  we  have 

w  +  o  :  b  : :  a  :  c ;  or,  cw  +  <?2  =  ab ;     ( 1 ) 

But,  as  EO  and  J.1?  are  two  chords  that  intersect  each 
other  in  a  circle,  we  have 

cw  =  xy,     (Th.  IT,  B.  III). 
Therefore,  xy  +  c*  =  ab.     ( 2 ) 

But,  as  (7D  bisects  the  vertical  augle,  v;  e  have 

a  :  b  ::  x  :  y,     (Th.  24,  B.  II). 

Or, 

Hence, 

And, 

Now,  as  x  and  #  are  determined,  the  base  is  deter- 
mined. 

Remark.  —  Ot serve  that  equation  (2)  is  Theorem  20,  Book  III 
20* 


x  = 
+  <?'  = 

'  b' 

■  ab; 

(3) 
or,  y  = 

a 

-\A- 

jfib 

a 

X- 

mW»- 

a 

234 


GEOMETRY. 


PROBLEM    IV. 

To  determine  a  triangle,  from  the  base,  the  line  bisecting 
the  vertical  angle,  and  the  diameter  of  the  circumscribing  circle. 

Describe  the  circle  on  the  given 
diameter,  AB,  and  divide  it  into  two 
parts,  in  the  point  D,  so  that  AD  x 
DB  shall  be  equal  to  the  square  of 
one  half  the  given  base,  (Th.  17,  B.  III). 

Through   D  draw  EDG,   at  right 
angles  to  AB,  and  EG  will  be  the  given  base  of  the 
triangle. 

Put        AB  —  n,  DB  =  m,  AB  =  d,  DG  =  b. 


Then, 


n  -f  m  =  d,  and 


nm 


and  these  two  equations  will  determine  n  and  m ;  there- 
fore, we  shall  consider  n  and  m  as  known. 

Now,  suppose  EEG  to  be  the  required  A;  and  draw 
RIB  and  HA.  The  two  A's,  ABE,  DBI,  are  equian- 
gular ;  and,  therefore,  we  have 

AB  :  EB  : :  IB  :  DB. 

But  EI  is  a  given  line,  that  we  will  represent  by  c ; 
and  if  we  put  IB  —  w,  we  shall  have  EB  =  c  +  w;  then 
the  above  proportion  becomes, 

d  :  c  +  w  : :  w  :  m. 

Now,  w  can  be  determined  by  a  quadratic  equation ; 
and,  therefore,  IB  is  a  known  line. 

In  the  right-angled  A  DBI,  the  hypotenuse  IB,  and 
the  base  DB,  are  known ;  therefore,  DI  is  known,  (Th. 
39,  B.  I) ;  and  if  DI  is  known,  EI  and  IG  are  known. 

Lastly,  let  EH=  x,  EG  =  y,  and  put  EI=  p,  and  IG 
-?• 

Then,  by  Theorem  20,  Book  III,  pq  +  <?'  =  org     ( 1 ) 

But,  x  :  g  ::  p  :  q  (Th. 24,  B.  II) 


BOOK  V.II 


235 


Or, 


x=*l 


(2) 


Now,  from  equations  ( 1 )  and  ( 2 )  we  can  determine  x 
andy,  the  sides  of  the  A ;  and  thus  the  determination  has 
been  attained,  carefully  and  easily,  step  by  step. 


PROBLEM   V. 

Three  equal  circles  touch  each  other  externally,  and  thus 
inclose  one  acre  of  ground;  what  is  the  diameter  in  rods  of 
each  of  these  circles  f 

Draw  three  equal  circles  to  touch  each  other  exter- 
nally, and  join  the  three  centers,  thus  forming  a  triangle. 
The  lines  joining  the  centers  will  pass 
through  the  points  of  contact,  (Th.  7, 
B.IH). 

Let  R  represent  the  radius  of  these 
equal  circles;  then  it  is  obvious  that 
each  side  of  this  A  is  equal  to  2R. 
The  triangle  is  therefore  equilateral, 
and  it  incloses  the  given  area,  and  three  equal  sectors. 

As  the  angle  of  each  sector  is  one  third  of  two  right 
angles,  the  three  sectors  are,  together,  equal  to  a  semi- 
circle ;  but  the  area  of  a  semi-circle,  whose  radius  is  R,  is 

AT  J?2 

expressed  by  — ;   and  the  area  of  the  whole  triangle 

irR2 

must  be  — -  -f  160 ;  but  the  area  of  the  A  is  also  equal  to 

R   multiplied  by  the   perpendicular  altitude,  which   is 
fiv/3. 


Therefore, 
Or, 


jRV3  =  ^  +  160. 


22»(2%/3  —  *)  =  320. 
320 


320 


-  992.248. 


2  v/3  —  3.1415926      0.3225 
Hence,  R  =  31.48  +  rods,  for  the  required  result 


286  GEOMETRY. 

Problem  VI.  —  In  a  right-angled  triangle,  having  given 
the  base  and  the  sum  of  the  perpendicular  and  hypotennse, 
to  find  these  two  sides, 

Prob.  VII. — Given,  the  base  and  altitude  of  a  triangle,  to 
divide  it  into  three  equal  parts,  by  lines  parallel  to  the  base. 

Prob.  VIII. — In  any  equilateral  A,  given  the  length  of 
the  three  perpendiculars  drawn  from  any  point  within,  to  the 
three  sides,  to  determine  the  sides* 

Prob.  IX. — In  a  right-angled  triangle,  having  given  the 
base,  ( 3 ),  and  the  difference  between  the  hypotenuse  and  per- 
pendicular,  ( 1 ),  to  find  both  these  two  sides. 

Prob.  X. — In  a  right-angled  triangle,  having  given  the 
hypotenuse,  (5),  and  the  difference  between  the  base  and 
perpendicular,  ( 1 ),  to  determine  both  these  two  sides. 

Prob.  XI. — Having  given  the  area  of  a  rectangle  inscribed 
in  a  given  triangle,  to  determine,  the  sides  of  the  rectangle. 

Prob.  XII. — In  a  triangle,  having  given  the  ratio  of  the 
two  sides,  together  with  both  the  segments  of  the  base,  made 
by  a  perpendicular  from  the  vertical  angle,  to  determine  the 
sides  of  the  triangle. 

Prob.  XIH. — In  a  triangle,  having  given  the  base,  the 
sum  of  the  other  two  sides,  and  the  length  of  a  line  drawn 
from  the  vertical  angle  to  the  middle  of  the  base,  to  find  the 
sides  of  the  triangle. 

Prob.  XTV. — To  determine  a  right-angled  triangle,  having 
given  the  lengths  of  two  lines  drawn  from  the  acute  angles  to 
*he  middle  of  the  opposite  sides. 

Prob.  XV. — To  determine  a  right-angled  triangle,  having 
given  the  perimeter,  and  the  radius  of  the  inscribed  circle. 

Prob.  XVI. —  To  determine  a  triangle,  having  given  the 
base,  the  perpendicular,  and  the  ratio  of  the  two  sides. 

Prob.  XVII. —  To  determine  a  right-angled  triangle,  having 
given  the  hypotenuse,  and  the  side  of  the  inscribed  square. 


BOOK   VIII.  237 

Prob.  XV  III.  —  To  determine  the  radii  of  three  equal  cir- 
cles inscribed  in  a  given  circle,  and  tangent  to  each  other,  and 
also  to  the  circumference  of  the  given  circle. 

Prob.  XIX.  — In  a  right-angled  triangle,  having  given  the 
perimeter,  or  sum  of  all  the  sides,  and  the  perpendicular  let 
fall  from  the  right  angle  on  the  hypotenuse,  to  determine  the 
triangle  ;  that  is,  its  sides. 

Prob.  XX. — To  determine  a  right-angled  triangle,  having 
given  the  hypotenuse,  and  the  difference  of  two  lines  drawn 
from  the  two  acute  angles  to  the  center  of  the  inscribed  circle. 

Prob.  XXI.  —  To  determine  a  triangle,  having  given  the 
base,  the  perpendicular,  and  the  difference  of  the  two  other 
fades. 

Prob.  XXII.  —  To  determine  a  triangle,  having  given  the 
base,  the  perpendicular,  and  the  rectangle,  or  product  of  the 
two  sides. 

Prob.  XXIII. — To  determine  a  triangle,  having  given  the 
lengths  of  three  lines  drawn  from  the  three  angles  to  the  mid- 
dle of  the  opposite  sides. 

Prob.  XXTV.  —  In  a  triangle,  having  given  all  the  three 
Bides,  to  find  the  radius  of  the  inscribed  circle. 

Prob.  XXV. — To  determine  a  right-angled  triangle,  having 
given  the  side  of  the  inscribed  square,  and  the  radius  of  the 
inscribed  circle. 

Prob.  XXVI.  —  To  determine  a  triangle,  and  the  radius 
of  the  inscribed  circle,  having  given  the  lengths  of  three  lines 
drawn  from  the  three  angles  to  the  center  of  that  circle. 

Prob.  XXVTI.  —  To  determine  a  right -angled  triangle, 
having  given  the  hypotenuse,  and  the  radius  of  the  inscribed 
tircle. 

Prob.  XXVJLLL. — The  lengths  of  two  parallel  chords  on  the 
tame  side  of  the  center  being  given,  and  their  distance  apart, 
to  determine  the  radius  of  the  circle. 

Prob,  XXIX.  —  The  lengths  of  two  chords  in  the  same 


238  GEOMETRY. 

circle  being  given,  and  also  the  difference  of  their  distance 
from  the  center,  to  find  the  radius  of  the  circle. 

Prob.  XXX. — The  radius  of  a  circle  being  given,  and  also 
the  rectangle  of  the  segments  of  a  chord,  to  determine  the  dis- 
tance of  the  point  at  which  the  chord  is  divided,  from  the 
center. 

Prob.  XXXI. — If  each  of  the  two  equal  sides  of  an  isos- 
celes triangle  be  represented  by  a,  and  the  base  by  2b,  what 
will  be  the  value  of  the  radius  of  the  inscribed  circle  ? 

A         j,       b^a*  —  b* 

Ans.  R  = - — . 

a  +  b 

Prob.  XXXII.  —  From  a  point  without  a  circle  whose 
diameter  is  d,  a  line  equal  to  d  is  drawn,  terminating  in  the 
concave  arc,  and  this  line  is  bisected  at  the  first  point  in  which 
it  meets  the  circumference.  What  is  the  distance  of  the  point 
without  from  the  center  of  the  circle  ? 

It  is  not  deemed  necessary  to  multiply  problems  in  the 
application  of  algebra  to  geometry.  The  preceding  will 
be  a  sufficient  exercise  to  give  the  student  a  clear  con- 
ception of  the  nature  of  such  problems,  and  will  serve  as 
a  guide  for  the  solution  of  others  that  may  be  proposed 
to  him,  or  that  may  be  invented  by  his  own  ingenuity. 

MISCELLANEOUS   PROPOSITIONS. 

We  shall  conclude  this  book,  and  the  subject  of  Geom- 
etry, by  offering  the  following  propositions,  —  some  the- 
orems, others  problems,  and  some  a  combination  of  both, 
— not  only  for  the  purpose  of  impressing,  by  application, 
the  geometrical  principles  which  have  now  been  estab- 
lished, but  for  the  not  less  important  purpose  of  culti- 
vating the  power  of  independent  investigation. 

After  one  or  two  propositions  in  which  the  beginner 
will  be  assisted  in  the  analysis  and  construction,  we  shall 
leave  him  to  his  own  resources,  with  the  caution  that  a 


BOOK    VIII. 


239 


patient  consideration  of  all  the  conditions  in  each  case, 
and  not  mere  trial  operation,  is  the  only  process  by  which 
he  can  hope  to  reach  the  desired  result. 

1.  From  two  given  points,  to  draw  two  equal  straight 
lines,  which  shall  meet  in  the  same  point  in  a  given 
straight  line. 

Let  A  and  B  be  the  given  points, 
and  CD  the  given  straight  line.  Pro- 
duce the  perpendicular  to  the  straight 
line  AB  at  its  middle  point,  until  it 
meets  CD  in  G.  It  is  then  easily 
proved  that  G  is  the  point  in  CD  in 
which  the  equal  lines  from  A  and 
B  must  meet.  That  is,  that  AG 
=  BG. 

If  the  points  A  and  B  were  on 
opposite  sides  of  CD,  the  directions 
for  the  construction  would  be  the 
same,  and  we  should  have  this  fig- 
ure; but  the  reasoning  by  which 
we  prove  AG  =  BG  would  be  un- 
changed. 

2.  From  two  given  points  on  the  same  side  of  a  given 
straight  line,  to  draw  two  straight  lines  which  shall  meet 
in  the  given  line,  and  make  equal  angles  with  it. 

Let  CD  be  the  given  line,  and 
A  and  B  the  given  points. 

From  B  draw  BE  perpendicular 
to  CD,  and  produce  the  perpen- 
dicular to  F,  making  EF  equal  to 
BE)  then  draw  AF,  and  from  the 
point  G,  in  which  it  intersects 
CD,  draw  GB.  Now,  [__BGE=> 
\_EGF  =\_AGC.  Hence,  the 
angles  B  GD  and  A  G  C  are  equal, 
and  the  lines  AG  and  BG  meet 

in  a  common  point  in  the  line  CD,  and  made  equal  angles  witk 
that  line. 


240  GEOMETRY. 

3.  If,  from  a  point  without  a  circle,  two  straight  lines 
be  drawn  to  the  concave  part  of  the  circumference,  making 
equal  angles  with  the  line  joining  the  same  point  and  the 
center,  the  parts  of  these  lines  which  are  intercepted  within 
the  circle,  are  equal. 

4.  If  a  circle  be  described  on  the  radius  of  another  circle, 
any  straight  line  drawn  from  the  point  where  they  meet, 
to  the  outer  circumference,  is  bisected  by  the  interior  one. 

5.  From  two  given  points  on  the  same  side  of  a  line 
given  in  position,  to  draw  two  straight  lines  which  shall 
contain  a  given  angle,  and  be  terminated  in  that  line. 

6.  If,  from  any  point  without  a  circle,  lines  be  drawn 
touching  the  circle,  the  angle  contained  by  the  tangents  is 
double  the  angle  contained  by  the  line  joining  the  points 
of  contact  and  the  diameter  drawn  through  one  of  them. 

7.  If,  from  any  two  points  in  the  circumference  of  a 
circle,  there  be  drawn  two  straight  lines  to  a  point  in  a 
tangent  to  that  circle,  they  will  make  the  greatest  angle 
when  drawn  to  the  point  of  contact. 

8.  From  a  given  point  within  a  giv^n  circle,  to  draw  a 
straight  line  which  shall  make,  with  the  circumference, 
an  angle,  less  than  any  angle  made  by  any  other  line 
drawn  from  that  point. 

9.  If  two  circles  cut  each  other,  the  greatest  line  that 
can  be  drawn  through  either  point  of  intersection,  is  that 
which  is  parallel  to  the  line  joining  their  centers. 

10.  If,  from  any  point  within  an  equilateral  triangle, 
perpendiculars  be  drawn  to  the  sides,  their  sum  is  equal 
to  a  perpendicular  drawn  from  any  of  the  angles  to  the 
opposite  side. 

11.  If  the  points  of  bisection  of  the  sides  of  a  given  tri- 
angle be  joined,  the  triangle  so  formed  will  be  one  fourth 
of  the  given  triangle. 

12.  The  difference  of  the  angles  at  the  base  of  any  tri- 
angle, is  double  the  angle  contained  by  a  line  drawn  from 
the  vertex  perpendicular  to  the  base,  and  another  bisect- 
ing  the  angle  at  the  vertex. 


BOOK   VIII.  241 

13.  If,  from  the  three  angles  of  a  triangle,  lines  be 
*rawn  to  the  points  of  bisection  of  the  opposite  sides, 
tfiese  lines  intersect  each  other  in  the  same  point. 

14.  The  three  straight  lines  which  bisect  the  three 
!  angles  of  a  triangle,  meet  in  the  same  point. 

15.  The  two  triangles,  formed  by  drawing  straight 
lines  from  any  point  within  a  parallelogram  to  the  ex- 
tremities of  two  opposite  sides,  are,  together,  one  half  the 
parallelogram. 

16.  The  figure  formed  by  joining  the  points  of  bisection 
of  the  sides  of  a  trapezium,  is  a  parallelogram. 

17.  If  squares  be  described  on  three  sides  of  a  right- 
angled  triangle,  and  the  extremities  of  the  adjacent  sides 
be  joined,  the  triangles  so  formed  are  equivalent  to  the 
given  triangle,  and  to  each  other. 

18.  If  squares  be  described  on  the  hypotenuse  and  sides 
of  a  right-angled  triangle,  and  the  extremities  of  the  sides 
of  the  former,  and  the  adjacent  sides  of  the  others,  be 
joined,  the  sum  of  the  squares  of  the  lines  joining  them 
will  be  equal  to  five  times  the  square  of  the  hypotenuse. 

19.  The  vertical  angle  of  an  oblique-angled  triangle 
inscribed  in  a  circle,  is  greater  or  less  than  a  right  angle, 
by  the  angle  contained  between  the  base  and  the  diam- 
eter drawn  from  the  extremity  of  .the  base. 

20.  If  the  base  of  any  triangle  be  bisected  by  the  diam- 
eter of  its  circumscribing  circle,  and,  from  the  extremity 
of  that  diameter,  a  perpendicular  be  let  fall  upon  the 
longer  side,  it  will  divide  that  side  into  segments,  one  of 
which  will  be.  equal  to  one  half  the  sum,  and  the  other  to 
one  half  the  difference, of  the  sides. 

21.  A  straight  line  drawn  from  the  vertex  of  an  equi- 
lateral triangle  inscribed  in  a  circle,  to  any  point  in  the 
opposite  circumference,  is  equal  to  the  sum  of  the  two  lines 
which  are  drawn  from  the  extremities  of  the  base  to  the 
same  point. 

22.  The  straight  line  bisecting  any  angle  of  a  triangle 

21  Q 


242  GEOMETRY. 

inscribed  in  a  given  circle,  cuts  the  circumference  in  a 
point  which  is  equi-distant  from  the  extremities  of  the 
side  opposite  to  the  bisected  angle,  and  from  the  center 
of  a  circle  inscribed  in  the  triangle. 

23.  If,  from  the  center  of  a  circle,  a  line  be  drawn  to 
any  point  in  the  chord  of  an  arc,  the  square  of  that  line, 
together  with  the  rectangle  contained  by  the  segments 
of  the  chord,  will  be  equal  to  the  square  described  on  the 
radius. 

24.  If  two  points  be  taken  in  the  diameter  of  a  circle, 
equidistant  from  the  center,  the  sum  of  the  squares  of  the 
two  lines  drawn  from  these  points  to  any  point  in  the  cir- 
cumference, will  be  always  the  same. 

25.  If,  on  the  diameter  of  a  semicircle,  two  equal  circles 
be  described,  and  in  the  space  included  by  the  three  cir- 
cumferences, a  circle  be  inscribed,  its  diameter  will  be  % 
the  diameter  of  either  of  the  equal  circles. 

26.  If  a  perpendicular  be  drawn  from  the  vertical  angle 
of  any  triangle  to  the  base,  the  difference  of  the  squares 
of  the  sides  is  equal  to  the  difference  of  the  squares  of 
the  segments  of  the  base. 

27.  The  square  described  on  the  side  of  an  equilateral 
triangle,  is  equal  to  three  times  the  square  of  the  radius 
of  the  circumscribing  circle. 

28.  The  sum  of  the  sides  of  an  isosceles  triangle  is  less 
than  the  sum  of  the  sides  of  any  other  triangie  on  the  same 
base  and  between  the  same  parallels. 

29.  In  any  triangle,  given  one  angle,  a  side  adjacent  to 
the  given  angle,  and  the  difference  of  the  other  two  sides, 
to  construct  the  triangle. 

30.  In  any  triangle,  given  the  base,  the  sum  of  the 
other  two  sides,  and  the  angle  opposite  the  base,  to  con- 
struct the  triangle. 

31.  In  any  triangle,  given  the  base,  the  angle  opposite 
io  the  base,  and  the  difference  of  the  other  two  sides,  to 
instruct  the  triangle. 


BOOK   IX.  243 


BOOK    IX. 


SPHERICAL    GEOMETRY. 


DEFINITIONS. 

1.  Spherical  Geometry  has  for  its  object  the  investiga- 
tion of  the  properties,  and  of  the  relations  to  each  other, 
of  the  portions  of  the  surface  of  a  sphere  which  are 
bounded  by  the  arcs  of  its  great  circles. 

2.  A  Spherical  Polygon  is  a  portion  of  the  surface  of  a 
sphere  bounded  by  three  or  more  arcs  of  great  circles,  called 
the  sides  of  the  polygon. 

3.  The  Angles  of  a  spherical  polygon  are  the  angles 
formed  by  the  bounding  arcs,  and  are  the  same  as  the 
angles  formed  by  the  planes  of  these  arcs. 

4.  A  Spherical  Triangle  is  a  spherical  polygon  having 
but  three  sides,  each  of  which  is  less  than  a  semi-circum- 
ference. 

5.  A  Lime  is  a  portion  of  the  surface  of  a  sphere  in- 
cluded between  two  great  semi-circumferences  having  a 
common  diameter. 

6.  A  Spherical  Wedge,  or  TTngula,  is  a  portion  of  the 
Bolid  sphere  included  between  two  great  semi-circles  having 
a  common  diameter 


244  GEOMETRY. 

7.  A  Spherical  Pyramid  is  a  portion  of  a  sphere  bounded 
by  the  faces  of  a  solid  angle  having  its  vertex  at  the 
center,  and  the  spherical  polygon  which  these  faces  inter- 
cept on  the  surface.  This  spherical  polygon  is  called  the 
base  of  the  pyramid. 

8.  The  Axis  of  a  great  circle  of  a  sphere  is  that  diameter 
of  the  sphere  which  is  perpendicular  to  the  plane  of  the 
circle.  This  diameter  is  also  the  axis  of  all  small  circles 
parallel  to  the  great  circle. 

9.  A  Pole  of  a  circle  of  a  sphere  is  a  point  on  the  sur- 
face of  the  sphere  equally  distant  from  every  point  in  the 
circumference  of  the  circle. 

10.  Supplemental,  or  Polar  Triangles,  are  two  triangles  on 
a  sphere,  so  related  that  the  vertices  of  the  angles  of 
either  triangle  are  the  poles  of  the  sides  of  the  other. 

PKOPOSITION   I. 

Any  two  sides  of  a  spherical  triangle  are  together  greater 
than  the  third  side. 

Let  AB,  AC,  and  BC,  be  the  three 
sides  of  the  triangle,  and  D  the  center 
of  the  sphere. 

The  angles  of  the  planes  that  form 
the  solid  angle  at  D,  are  measured  by 
the  arcs  AB,  AG,  snidBC.  But  any 
two  of  these  angles  are  together  greater 
than  the  third  angle,  (Th.  18,  B.  VI).  Therefore,  any  two 
sides  of  the  triangle  are,  together,  greater  than  *  ue  third  side. 

Hence  the  proposition. 

PROPOSITION   II. 

The  sum  of  the  three  sides  of  any  spherical  triangle  is  less 
than  the  circumference  of  a  great  circle. 

Let  ABC  be  a  spherical  triangle ;  the  two  sides,  AB 
and  AC,  produced,  will  meet  at  the  point  which  is  diame- 
trically opposite  to  Af  and  the  arcs,  ABB  and  ACB  are 


BOOK    IX 


245 


together  equal  to  a  great  circle.  But, 
by  the  last  proposition,  BC  is  less 
than  the  two  arcs,  BD  and  DC  There- 
fore, AB  +  BO  +  AC,  is  less  than 
ABD  +  ACD;  that  is,  less  than  a 
great  circle. 

Hence  the  proposition. 

PROPOSITION    III. 

The  extremities  of  the  axis  of  a  great  circle  of  a  sphere 
are  the  poles  of  the  great  circle,  and  these  points  are  also 
the  poles  of  all  small  circles  parallel  to  the  great  circle. 

Let  0  be  the  center  of 
the  sphere,  and  BD  the 
axis  of  the  great  circle, 
Cm  Am" ;  then  will  B  and 
D,  the  extremities  of  the 
axis,  be  the  poles  of  the 
circle,  and  also  the  poles 
of  any  parallel  small  cir- 
cle, as  FnE. 

For,  since  BD  is  per- 
pendicular to  the  plane 
of  the  circle,  Cm  Am",  it 

is  perpendicular  to  the  lines  OA,  0m',  Om",  etc.,  passing 
through  its  foot  in  the  plane,  (Def.  2,  B.  VI);  hence,  all 
the  arcs,  Bm,  Bm',  etc.,  are  quadrants,  as  are  also  the 
arcs  Dm,  Dm',  etc.  The  points  B  and  D  are,  therefore, 
each  equally  distant  from  all  the  points  in  the  circumfer- 
ence, Cm  Am" ;  hence,  (Def.  9),  they  are  its  poles. 

Again,  since  the  radius,  OB,  is  perpendicular  to  the 
plane  of  the  circle,  Cm  Am",  it  is  also  perpendicular  to 
the  plane  of  the  parallel  small  circle,  FnE,  and  passes 
through  its  center,  0'.  Now,  the  chords  of  the  arcs,  BF, 
Bn,  BE,  etc.,  being  obiique  lines,  meeting  the  plane  of 
the  small  circle  a4-  oqual  distances  from  the  foot  of  the 


o" -- 

\e 

/  IS 

— * — Z^- 

hi 

m» 

[.-""'-""/ 

/ 

■"---», 

\         iJtr 

0 

\j)f 

246  GEOMETRY. 

perpendicular,  BO',  are  all  equal,  (Th.  4,  B.  VI);  hence, 
the  arcs  themselves  are  equal,  and  B  is  one  pole  of  the 
circle,  FnE.  In  like  manner  we  prove  the  arcs,  BF,  Dn, 
BE,  etc.,  equal,  and  therefore  D  is  the  other  pole  of  the 
same  circle. 

Hence  the  proposition,  etc. 

Cor.  1.  A  point  on  the  surface  of  a  sphere  at  the  distance 
of  a  quadrant  from  two  points  in  the  arc  of  a  great  circle,  not 
at  the  extremities  of  a  diameter,  is  a  pole  of  that  arc. 

For,  if  the  arcs,  Bm,  Bmr,  are  each  quadrants,  the  angles, 
BOm  and  BOmf,  are  each  right  angles;  and  hence,  BO 
is  perpendicular  to  the  plane  of  the  lines,  Om  and  0mf, 
which  is  the  plane  of  the  arc,  mm';  B  is  therefore  the 
pole  of  this  arc. 

Cor.  2.  The  angle  included  between  the  arc  of  a  great  circle 
and  the  arc  of  another  great  circle,  connecting  any  of  its  points 
with  the  pole,  is  a  right  angle. 

For,  since  the  radius,  BO,  is  perpendicular  to  the  plane 
of  the  circle,  Cm  Am",  every  plane  passed  through  this 
radius  is  perpendicular  to  the  plane  of  the  circle ;  hence, 
the  plane  of  the  arc  Bm  is  perpendicular  to  that  of  the 
arc  (7m;  and  the  angle  of  the  arcs  is  that  of  their  planes. 

PKOPOSITION   IV. 

The  angle  formed  by  two  arcs  of  great  circles  which  inter- 
sect each  other,  is  equal  to  the  angle  included  between  the  tan- 
gents to  these  arcs  at  their  point  of  intersection,  and  is  meas- 
ured by  that  arc  of  a  great  circle  whose  pole  is  the  vertex  of 
the  angle,  and  which  is  limited  by  the  sides  of  the  angle  or 
the  sides  produced. 

Let  AM  and  AN  be  two  arcs  intersecting  at  the 
point  A,  and  let  AE  and  AF  be  the  tangents  to  these 
arcs  at  this  point.  Take  AC  and  AD,  each  quadrants, 
and  draw  the  arc  CD,  of  which  A  is  the  pole,  and  OQ 
and  OB  are  the  radii. 


BOOK    IX, 


247 


Now,  since  the  planes  of  the  arcs  intersect  in  the  radius 
OA,  and  AE  is  a  tangent  to  one  arc,  and  AF  a  tangent 
to  the  other,  at  the  common  point  'A,  A 
these  tangents  form  with  each  other  an 
angle  which  is  the  measure  of  the  angle 
of  the  planes  of  the  arcs ;  but  the  angle 
of  the  planes  of  the  arcs  is  taken  as  the 
angle  included  by  the  arcs,  (Def.  3). 

Again,  because  the  arcs,  AQ  and  AD, 
are  each  quadrants,  the  angles,  A  00, 
AOD,  are  right  angles ;  hence  the  radii, 
OC  and  OD,  which  lie,  one  in  one  face, 
and  the  other  in  the  other  face,  of  the 
diedral  angle  formed  by  the  planes  of  the  arcs,  are 
perpendicular  to  the  common  intersection  of  these  faces 
at  the  same  point.  The  angle,  OOD,  is  therefore  the 
angle  of  the  planes,  and  consequently  the  angle  of  the 
arcs  ;  but  the  angle  OOD  is  measured  by  the  arc  OD. 

Hence  the  proposition. 

Oor.  1.  Since  the  angles  included  between  the  arcs  of 
great  circles  on  a  sphere,  are  measured  by  other  arcs  of 
great  circles  of  the  same  sphere,  we  may  compare  such 
angles  with  each  other,  and  construct  angles  equal  to 
other  angles,  by  processes  which  do  not  differ  in  principle 
from  those  by  which  plane  angles  are  compared  and  con- 
structed. 

Oor.  2.  Two  arcs  of  great  circles  will  form,  by  their  in- 
tersection, four  angles,  the  opposite  or  vertical  ones  of 
which  will  be  equal,  as  in  the  case  of  the  angles  formed 
by  the  intersection  of  straight  lines,  (Th.  4,  B.  I). 


PROPOSITION  V. 

The  surface  of  a  hemisphere  may  be  divided  into  three  right- 
angled  and  four  quadrantal  triangles,  and  one  of  these  right- 
angled  triangles  will  be  so  related  to  the  other  two,  that  two 
of  its  sides  and  one  of  its  angles  will  be  complemental  to  the 


248  GEOMETRY. 

Bides  of  one  of  them,  and  two  of  its  sides  supplemental  to  two 
of  the  side.s  of  the  other. 

Let  ABO  be  a  right-angled  spherical  triangle,  right 
angled  at  B. 

Produce  the  sides,  AB  and  AC,  and 
they  will  meet  at  A',  the  opposite 
point  on  the  sphere.  Produce  BO, 
both  ways,  90°  from  the  point  B,  to 
P  and  P',  which  are,  therefore,  poles 
to  the  arc  AB,  (Prop.  3).  Through 
A,  P,  and  the  center  of  the  sphere, 
pass  a  plane,  cutting  the  sphere  into 
two  equal  parts,  forming  a  great  circle  on  the  sphere, 
which  great  circle  will  be  represented  by  the  circle 
PAP' A'  in  the  figure.  At  right  angles  to  this  plane, 
pass  another  plane,  cutting  the  sphere  into  two  equal 
parts ;  this  great  circle  is  represented  in  the  figure  by  the 
straight  line,  POP'.  A  and  A'  are  the  poles  to  the  great 
circle,  POP' ;  and  P  and  Pf  are  the  poles  to  the  great 
circle,  ABAf. 

!N"ow,  OPB  is  a  spherical  triangle,  right-angled  at  B, 
and  its  sides  OP  and  OB  are  complemental  respectively 
to  the  sides  BO  and  A 0  of  the  A  ABO,  and  its  side  PD 
is  complemental  to  the  arc  DO,  which  measures  the 
[_BAO  of  the  same  triangle.  Again,  the  A  A' BO  is  right- 
angled  at  B,  and  its  sides  A'O,  A'B,  are  supplemental 
respectively  to  the  sides  AO,  AB,  of  the  A  ABO.  There- 
fore, the  three  right-angled  A's,  ABO,  OPB,  and  A'BO, 
have  the  required  relations.  In  the  A  AOP,  the  side  AP 
is  a  quadrant,  and  for  this  reason  the  A  is  called  a  quad- 
ran  tal  triangle.  So  also,  are  the  A's  A' OP,  AOP',  and 
P'OA',  quadrantal  triangles.     Hence  the  proposition. 

Scholium. — In  every  triangle  there  are  six  elements,  three  sides  and 
three  angles,  called  the  parts  of  the  triangle. 

Now,  if  ail  the  parts  of  the  triangle  ABC  are  known,  the  parts  of 
each  of  the  A's>  PCD  and  A'BC,  are  as  completely  known.  And 
when  the  parts  of  the  A  PCD  are  known,  the  parts  of  the  A  '8  A  C2 


book  ix.  249 

and  A'CP  are  also  known  ;  for,  the  side  PD  measures  each  of  the  | 's 

P^lCand  PA'C,  and  the  angle  CPD,  added* to  the  right  angle  A' PDt 

gives  the  | A' PC,  and  the  | CPA  is  supplemental  to  this.     Hence, 

the  solution  of  the  A  ABC  is  a  solution  of  the  two  right-angled  and 
four  quadrantal  A's?  which  together  with  it  make  up  the  surface  of 
the  hemisphere. 

PROPOSITION  VI. 

If  there  be  three  arcs  of  great  circles  whose  poles  are  the 
angular  points  of  a  spherical  triangle,  such  arcs,  if  produced, 
will  form  another  triangle,  whose  sides  will  be  supplemental 
to  the  angles  of  the  first  triangle,  and  the  sides  of  the  first 
triangle  will  be  supplemental  to  the  angles  of  the  second. 

Let  the  arcs  of  the  three  great  cir- 
cles be  GH,  PQ,  KL,  whose  poles  are 
respectively  A,  B,  and  0.  Produce  the 
three  arcs  until  they  meet  in  D,  E,  and 
F.  We  are  now  to  prove  that  E  is  the 
pole  of  the  arc  AO;  D  the  pole  of  the 
arc  BO;  F  the  pole  to  the  arc  AB. 
Also,  that  the  side  EF,  is  supplemental 
to  the  angle  A;  EB  to  the  angle  0; 
and  BF  to  the  angle  B;  and  also,  that  the  side  A  0  is 
supplemental  to  the  angle  E,  etc. 

A  pole  is  90°  from  any  point  in  the  circumference  of 
Us  great  circle ;  and,  therefore,  as  A  is  the  pole  of  the 
arc  Gff,  the  point  A  is  90°  from  the  point  E.  As  0  is 
the  pole  of  the  arc  LK,  0  is  90°  from  any  point  in 
that  arc ;  therefore,  0  is  90°  from  the  point  E ;  and 
E  being  fcO°  from  both  A  and  0,  it  is  the  pole  of  the  arc 
AC.  In  the  same  manner,  we  may  prove  that  B  is  the 
pole  of  BO,  and  F  the  pole  of  AB. 

Because  A  is  the  pole  of  the  arc  Git,  the  arc  GH 
measures  the  angle  A,  (Prop.  4);  for  a  similar  reason, 
PQ  measures  the  angle  B,  and  LK  measures  the  angle  0, 
Because  E  is  the  pole  of  the  arc    AO,     EH  =  90° 
Or,  EG  +GH=  90° 

For  a  like  reason,  FH  +  GH  =  90° 


250  GEOMETRY. 

Adding  these  two  equations,  and  observing  that  QJf 
=  A,  and  afterward  transposing  one  A,  we  have, 
EG  +  GIT  +  FIT  =  180°  —  A. 

Or,  UF=1S0°—  A  } 

In  like  manner,  jF2>  =  180°  —  B    \     («) 

And,  BE  =  180°  —  0  J 

But  the  arc  (180°  — A),  is  a  supplemental  arc  to  A,  by 
the  definition  of  arcs;  therefore,  the  three  sides  of  the 
triangle  BEF,  are  supplements  of  the  angles  A,  By  C,  of 
the  triangle  ABO. 

Again,  as  E  is  the  pole  of  the  arc  A  C,  the  whole  angle 
E  is  measured  by  the  whole  arc  LH. 

But,  AC  +  CE  =  90° 

Also,  AQ  +  AL  =  90° 

By  addition,  AQ+AC+QH  +  AL  =  180° 
By  transposition,  ^(7+  (7^+^^  =  180° —J.  O 
That  is,  £#,  or  E=  180°— ^L<7  } 

In  the  same  manner,  F  =■  180°— ^    >  (&) 

And,  i)=180°  — J5(7  J 

That  is,  the  sides  of  the  first  triangle  are  supplemental 
to  the  angles  of  the  second  triangle. 

PROPOSITION   Til. 

The  sum  of  the  three  angles  of  any  spherical  triangle,  is 
greater  than  two  right  angles,  and  less  than  six  right  angles. 

Add  equations  («),  of  the  last  proposition.  The  first 
member  of  the  equation  so  formed  will  be  the  sum  of 
the  three  sides  of  a  spherical  triangle,  which  sum  we 
may  designate  by  S.  The  second  member  will  be  6  right 
angles  (there  being  2  right  angles  in  each  180°)  less  the 
three  angles  A,  B,  and  O. 

That  is,  S  =  6  right  angles  —  (A  +  B  +  C) 

By  Prop.  2,  the  sum  S  is  less  than  4  right  angles; 


BOOK   IX.  251 

therefore,  to  it  add  s,  a  sufficient  quantity  to  make  4 
right  angles.     Then, 

4  right  angles  =  6  right  angles  —  (A  +  B  -f  C)  -f  * 

Drop  or  cancel  4  right  angles  from  both  members,  and 
transpose  (A  +  B  +  O). 

Then,  A  +  B  +  0  =  2  right  angles  +  a. 

That  is,  the  three  angles  of  a  spherical  triangle  make 
a  greater  sum  than  two  right  angles  by  the  indefinite 
quantity  *,  which  quantity  is  called  the  spherical  excess, 
and  is  greater  or  less  according  to  the  size  of  the  triangle. 

Again,  the  sum  of  the  angles  is  less  than  6  right  angles. 
There  are  but  three  angles  in  any  triangle,  and  each  one  of 
them  must  be  less  than  180°,  or  2  right  angles.  For,  an 
angle  is  the  inclination  of  two  lines  or  two  planes ;  and 
when  two  planes  incline  by  180°,  the  planes  are  parallel, 
or  are  in  one  and  the  same  plane ;  therefore,  as  neither 
angle  can  be  equal  to  2  right  angles,  the  three  can  never 
be  equal  to  6  right  angles. 

PROPOSITION   VIII. 

On  the  same  sphere,  or  on  equal  spheres,  triangles  which 
are  mutually  equilateral  are  also  mutually  equiangular  ;  and, 
conversely,  triangles  which  are  mutually  equiangular  are  also 
mutually  equilateral,  equal  sides  lying  opposite  equal  angles. 

First— -Let  ABO  and  DEF,  in 
which  AB  =  BE,  AO=  DF,  and 
BO  =  EF,  be  two  triangles  on 
the  sphere  whose  center  is  0; 
then  will  the  [_  A,  opposite  the 
side  BO,  in  the  first  triangle,  be 
equal  the  [_D,  opposite  the  equal 
side  EF,  in  the  second;  also 
L#  =  l    E,  andl_<?=t    F. 


252  GEOMETBY. 

For,  drawing  the  radii  to  the  vertices  of  the  angles  of 
these  triangles,  we  may  conceive  0  to  he  the  common 
vertex  of  two  triedral  angles,  one  of  which  is  hounded 
by  the  plane  angles  A  OB,  BOO,  and  A  00,  and  the  other 
by  the  plane  angles  DOE,  EOF,  and  J)OF.  But  the 
plane  angles  bounding  the  one  of  these  triedral  angles, 
are.  equal  to  the  plane  angles  bounding  the  other,  each 
to  each,  since  they  are  measured  by  the  equal  sides  of  the 
two  triangles.  The  planes  of  the  equal  arcs  in  the  two 
triangles  are  therefore  equally  inclined  to  each  other, 
(Th.  20,  B.  VI) ;  but  the  angles  included  between  the 
planes  of  the  arcs  are  equal  to  the  angles  formed  by  the 
arcs,  (Def.  -3). 

Hence  the  [_  A,  opposite  the  side  BO,  in  the  A  ABU, 
is  equal  to  the  [_  -A  opposite  the  equal  side  EF,  in  the 
other  triangle ;  and  for  a  similar  reason,  the  [__B=  l_E, 
and  the  l_0=l_F. 

Second. — If,  in  the  triangles  ABO  and  DEF,  being  on 
the  same  sphere  whose  center  is  0,  the  [_A  =  [_D,  the 
\_B  =  [_E,  and  the  L<7«  L^7;  tnen  will  the  side  AB, 
opposite  the  [__  0,  in  the  first,  be  equal  to  the  side  BE, 
opposite  the  equal  [__F,  in  the  second;  and  also  the  side 
A 0  equal  to  the  side  BF,  and  the  side  BO  equal  to  the 
side  EF. 

For,  conceive  two  triangles,  denoted  by  A'B'Of  and 
D'E'F',  supplemental  to  ABO  and  BEF,  to  be  formed; 
then  will  these  supplemental  triangles  be  mutually  equi- 
lateral, for  their  sides  are  measured  by  180°  less  the 
o]  posite  and  equal  angles  of  the  triangles  ABO  and 
DEF,  (Prop.  6) ;  and  being  mutually  equilateral,  they 
are,  as  proved  above,  mutually  equiangular.  But  the 
triangles  ABO  and  DEF  are  supplemental  to  the  tri- 
angles A'B'O'  and  D'E'F1 ';  and  their  sides  are  therefore 
measured  severally  by  180°  less  the  opposite  and  equal 
angles  of  the  triangles  A'B'O'  and  D'E'2",  (Prop.  6), 


book  ix.  253 

Hence  the  triangles  ABO  and  DEF,  which  are  mutually 
equiangular,  are  also  mutually  equilateral. 

Scholium. — With  the  three  arcs  of  great  circles,  AB,  AC,  and  BCt 
either  of  the  two  triangles,  ABC,  DEF,  may  be  formed ;  but  it  is  evi- 
dent that  these  two  triangles  cannot  be  made  to  coincide,  though  they 
are  both  mutually  equilateral  and  mutually  equiangular.  Spherical 
triangles  on  the  same  sphere,  or  on  equal  spheres,  in  which  the  sides 
and  angles  of  the  one  are  equal  to  the  sides  and  angles  of  the  other, 
each  to  each,  but  are  not  themselves  capable  of  superposition,  are 
called  symmetrical  triangles. 


PROPOSITION    IX. 

On  the  same  sphere,  or  on  equal  spheres,  triangles  having 
two  sides  of  the  one  equal  to  two  sides  of  the  other,  each  to 
each,  and  the  included  angles  equal,  have  their  remainmg 
sides  and  angles  equal. 

Let  ABO  and  DEF  be  two 

triangles,  in  which  AB  —  DE, 
AO  =  DF,  and  the  angle  A  = 
the  angle  D ;  then  will  the  side 
BO  be  equal  to  the  side  FE, 
the  L  B  =  the  [__E,  and  |_  G 

For,  if  DE  lies  on  the  same 
side  of  DjFthat  AB  does  of  AO,  the  two  triangles,  ABC 
and  DEF,  may  be  applied  the  one  to  the  other,  and  they 
may  be  proved  to  coincide,  as  in  the  case  of  plane  tri- 
angles. But,  if  DE  does  not  lie  on  the  same  side  of  DF 
that  AB  does  of  AO,  we  may  construct  the  triangle  which 
is  symmetrical  with  DEF;  and  this  symmetrical  triangle, 
when  applied  to  the  triangle  ABO,  will  exactly  coincide 
with  it.  But  the  triangle  DEF,  and  the  triangle  sym- 
metrical with  it,  are  not  only  mutually  equilateral,  but 
also  are  mutually  equiangular,  the  equal  angles  lying 
opposite  the  equal  sides,  (Prop.  8) ;  and  as  the  one  or  the 
other  will  coincide  with  the  triangle  ABO,  it  follows  that 


254  GEOMETRY 

the  triangles,  ABO  and  DBF,  are  either  absolutely  or 
symmetrically  equal. 

Oor.  On  the  same  sphere,  or  on  equal  spheres,  triangles 
having  two  angles  of  the  one  equal  to  two  angles  of  the  other, 
each  to  each,  and  the  included  sides  equal,  have  their  remain- 
ing sides  and  angles  equal. 

For,  if  \__A  =  L  A  L-#  -  L-^i  and  side  ^B  -  side 
DE,  the  triangle  DBF,  or  the  triangle  symmetrical  with 
it,  will  exactly  coincide  with  A  ABO,  when  applied  to  it 
as  in  the  case  of  plane  triangles ;  hence,  the  sides  and 
angles  of  the  one  will  be  equal  to  the  sides  and  angles 
of  the  other,  each  to  each. 


PKOPOSITION   X. 

In  an  isosceles  spherical  triangle,  the  angles  opposite  the 
equal  sides  are  equal. 

A 

Let  ABO  be  an  isosceles  spherical  tri- 
angle, in  which  AB  and  A  0  are  the  equal 
sides ;  then  will  [__  B  =  [__(?. 

For,  connect  the  vertex  A  with  D,  the  / 

middle  point  of  the  base,  by  the  arc  of  a  / 
great  circle,  thus  forming  the  two  mutu-  *f—~- 
ally  equilateral  triangles,  ABB  and  ADO. 
They  are  mutually  equilateral,  because  AD  is  common, 
BD  =  DO  by  construction,  and  AB=AO  by  supposition; 
hence  they  are  mutually  equiangular,  the  equal  angles 
being  opposite  the  equal  sides,  (Prop.  8).  The  angles  B 
and  0,  being  opposite  the  common  side  AD,  are  there- 
fore equal.  * 

Oor.  The  arc  of  a  great  circle  which  joins  the  vertex 
of  an  isosceles  spherical  triangle  with  the  middle  point  of 
the  base,  is  perpendicular  to  the  base,  and  bisects  the  ver- 
tical angle  of  the  triangle ;  and,  conversely,  the  arc  of  a 


BOOK    IX. 


255 


great  circle  which  bisects  the  vertical  angle  of  an  isosceles 
spherical  triangle,  is  perpendicular  to,  and  bisects  the 
base. 


PROPOSITION   XI. 

If  two  angles  of  a  spherical  triangle  are  equal,  the  opposite 
sides  are  also  equal,  and  the  triangle  is  isosceles. 
'  In  the  spherical  triangle,  ABO,  let  the  [_B  —  [_0;  then 
will  the  sides,  AB  and  A  0,  opposite  these  equal  angles, 
be  equal. 

For,  let  P  be  the  pole  of  the  base,  BO, 
and  draw  the  arcs  of  great  circles,  PB, 
PO;  these  arcs  will  be  quadrants,  and  at 
right  angles  to  BO,  (Cor.  2,  Prop.  3). 
Also,  produce  OA  and  BA  to  meet  PB 
and  PO,  in  the  points  E  and  F.  Wow, 
the  angles,  PBF  and  POE,  are  equal, 
because  the  first  is  equal  to  90°  less  the 
[__ABO,  and  the  second  is  equal  to  90° 
less  the  equal  \__AOB;  hence,  the  A's, 
PBFan&POE,  are  equal  in  all  their  parts, 
since  they  have  the  [_P  common,  the  [_PBF  =  \__POE, 
and  the  side  PB  equal  to  the  side  PO,  (Cor.,  Prop.  9). 
PE  is  therefore  equal  to  PF,  and  [_PEO=  \__PFB. 
1  Taking  the  equals  PF  and  PE,  from  the  equals  PO 
and  PB,  we  have  the  remainders,  FO  and  EB,  equal ; 
and,  from  180°,  taking  the  [_'s  PFB  and  PEO,  we  have 
the  remaining  [__'s,  AFO  and  AEB,  equal.  Hence,  the 
A's,  AFO  and  AEB,  have  two  angles  of  the  one  equal  to 
two  angles  of  the  other,  each  to  each,  and  the  included 
sides  equal;  the  remaining  sides  and  angles  are  therefore 
equal,  (Cor.,  Prop.  9).  Therefore,  A  0  is  equal  to  BA, 
and  the  A  ABO  is  isosceles. 

Cor.  An  equiangular  spherical  triangle  is  also  equilat- 
eral, and  the  converse. 


256  GEOMETRY. 

Behare. — In  this  demonstration,  the  pole  of  the  base,  BC,  is  sup- 
posed to  fall  without  the  triangle,  ABC.  The  same  figure  may  be  used 
for  the  case  in  which  the  pole  falls  within  the  triangle ;  the  modifi- 
cation the  demonstration  then  requires  is  so  slight  and  obvious,  that 
it  would  be  superfluous  to  suggest  it. 


PROPOSITION    XII. 

The  greater  of  two  sides  of  a  spherical  triangle  is  opposite 
the  greater  angle  ;  and,  conversely,  the  greater  of  two  angles 
of  a  spherical  triangle  is  opposite  the  greater  side. 

Let  ABC  be  a  spherical  triangle,  in  which  the  angle  A 
is  greater  than  the  angle  B ;  then  is  the  side  BO  greater 
than  the  side  A  C. 

Through  A  draw  the  arc  of  a 
great  circle,  AD,  making,  with  AB, 
the  angle  BAI)  equal  to  the  angle 
ABB.  The  triangle,  BAB,  is  isos- 
celes, and  DA  =  DB,  (Prop.  11). 

In  the  a  ACD,  CD+AD>AC, 
(Prop.  1.);  or,  substituting  for  AD  its  equal  DB,  we  have, 
CD  +  DB>  AC. 

If  in  the  above  inequality  we  now  substitute  CB  for 
CD-\-DB,  it  becomes  CB  >  CA. 

Conversely ;  if  the  side  CB  be  greater  than  the  side  CA, 
then  is  the  \_A  >  th«  [__i?.  For,  if  the  [_A  is  not  greater 
than  the  [_B,  it  is  either  equal  to  it,  or  less  than  it.  The 
[__A  is  not  equal  to  the  [_B ;  for  if  it  were,  the  triangle 
would  be  isosceles,  and  CB  would  be  equal  to  CA,  which 
is  contrary  to  the  hypothesis.  The  [_A  is  not  less  than 
the  [_B;  for  if  it  were,  the  side  CB  would  be  less  than  the 
side  CA,  by  the  first  part  of  the  proposition,  which  is  also 
contrary  to  the  hypothesis ;  hence,  the  [_A  must  be  greater 
than  the  L^- 


book  ix.  257 

PROPOSITION    XIII. 
Two  symmetrical  spherical  triangles  are  equal  in  area 

Let  ABO  and  BEF  be  two  A's  on  the  same  sphere, 
having  the  sides  and  angles  of  the  one  equal  to  the  sides 
and  angles  of  the  other,  each  to 
each,  the  triangles  themselves 
not  admitting  of  superposition. 
It  is  to  be  proved  that  these 
A's  have  equal  areas. 

Let  P  be  the  pole  of  a  small 
circle  passing  through  the  three 
points,  ABO,  and  connect  P 
with  each  of  the  points,  A,  B, 

and  0,  by  arcs  of  great  circles.  Next,  through  E  draw 
the  arc  of  a  great  circle,  EP',  making  the  angle  DEP' 
equal  to  the  angle  ABP.  Take  EP'  =  BP,  and  draw 
the  arcs  of  great  circles,  P'B,  P'F. 

The  A's,  ABP  and  BEP',  are  equal  in  all  their  parts, 
because  AB=BE,  BP=EP',  and  the  \__ABP=[_BEP', 
(Prop.  9).  Taking  from  the  \__ABO  the  \__ABP,  and 
from  the  \__DEF  the  \_BEP' ,  we  have  the  remaining 
angles,  PBO  and  P'EF,  equal;  and  therefore  the  A's, 
BOP  and  EFP1,  are  also  equal  in  all  their  parts. 

Now,  since  the  a's,  ABP  and  DEP',  are  isosceles,  they 
will  coincide  when  applied,  as  will  also  the  A's,  BOP 
and  EFP1,  for  the  same  reason.  The  polygonal  areas, 
ABQP  and  BEFP1,  are  therefore  equivalent.  If  from 
the  first  we  take  the  isosceles  triangle,  PAO,  and  from  the 
second  the  equal  isosceles  triangle,  P'BF,  the  remainders, 
or  the  triangle*  ABO  and  BEF,  will  be  equivalent. 

Remark.  —  It  is  assumed  in  this  demonstration  that  the  pole  P  falls 
without  the  triangle.  Were  it  to  fall  within,  instead  of  without,  no 
other  change  in  the  above  process  would  be  required  than  to  add  the 
isosceles  triangles,  PAC,  P/DF,  to  the  polygonal  areas,  to  get  the 
areas  of  the  triangles,  ABC,  DEF. 


258  GEOMETBY. 

Cor.  Two  spherical  triangles  on  the  same  sphere,  or  on 
equal  spheres,  will  be  equivalent  —  1st,  when  they  are 
mutually  equilateral;  —  2d,  when  they  are  mutually  equi- 
angular ;  —  3d,  when  two  sides  of  the  one  are  equal  to 
two  sideb  of  the  other,  each  to  each,  and  the  included 
angles  are  equal ;  —  4th,  when  two  angles  of  the  one  are 
equal  to  two  angles  of  the  other,  each  to  each,  and  the 
included  aides  are  equal. 

PKOPOSITION   XIV. 

If  two  arcs  of  great  circles  intersect  each  otlier  on  the  sur- 
face of  a  hemisphere,  the  sum  of  either  two  of  the  opposite  tri- 
angles thus  formed  will  he  equivalent  to  a  lune  whose  angle  is 
the  corresponding  angle  formed  by  the  arcs. 

Let  the  great  circle,  AEBC,  be  the  base  of  a  hemi- 
sphere, on  the  surface  of  which  the  great  semi-circumfer- 
ences, BBA  and  CBE,  inter- 
sect each  other  at  B ;  then  will 
the  sum  of  the  opposite  tri- 
angles, BBC  and  BAB,  be 
equivalent  to  the  lune  whose 
angle  is  BBC;  and  the  sum 
of  the  opposite  triangles, 
CBA  and  BBJE,  will  be  equiv- 
alent to  the  lune  whose  angle 
is  CBA. 

Produce  the  arcs,  BB A  and 
CBE,  until  they  intersect  on  the  opposite  hemisphere  at  H\ 
then,  since  CBE  and  BEE  are  both  semi-circumferences 
of  a  great  circle,  they  are  equal*.  Taking  from  each  the 
common  part  BE,  we  have  CB  =  HE.  In  the  same  way 
we  prove  BB  =  HA,  and  AE  =  BC.  The  two  triangles, 
BBC  and  HAE,  are  therefore  mutually  equilateral,  and 
hence  they  are  equivalent,  (Prop.  13).  But  the  two  tri- 
angles, BAE  and  ABEf  together,  make  up  the  lune 


BOOK    IX. 


259 


BEHAB\  hence  the  sum  of  the  a's,  BBO  and  ABE,  is 
equivalent  to  the  same  lune. 

By  the  same  course  of  reasoning,  we  prove  that  the 
sum  of  the  opposite  A's,  BAQ  and  BBE,  is  equivalent 
to  the  lune  BOHAB,  whose  angle  is  ABO. 


PROPOSITION   XV. 

The  surface  of  a  lune  is  to  the  whole  surface  of  the  sphere, 
as  the  angle  of  the  lune  is  to  four  right  angles ;  or,  as  the  arc 
which  measures  that  angle  is  to  the  circumference  of  a  great 
circle. 

I^ztABFCA  be  a  lune  on  the 
surface  of  a  sphere,  and  BCE 
an  arc  of  a  great  circle,  whose 
poles  are  A  and  F,  the  vertices 
of  the  angles  of  the  lune.  The 
arc,  BO,  will  then  measure  the 
angles  of  the  lune.  Take  any 
arc,  as  BB,  that  will  be  con- 
tained an  exact  number  of  times 
in  BO,  and  in  the  whole  circum- 
ference, BOEB,  and,  beginning  at  B,  divide  the  arc  and 
the  circumference  into  parts  equal  to  BB,  and  join  the 
points  of  division  and  the  poles,  by  arcs  of  great  circles. 
We  shall  thus  divide  the  whole  surface  of  the  sphere 
into  a  number  of  equal  lunes.  Now,  if  the  arc  BO  con- 
tains the  arc  BB  m  times,  and  the  whole  circumference 
contains  this  arc  n  times,  the  surface  of  the  lune  will 
oontain  m  of  these  partial  lunes,  and  the  surface  of  the 
sphere  will  contain  n  of  the  same ;  and  we  shall  have, 
Surf,  lune  :  surf,  sphere  : :  m  :  n. 

But,     m  :  n  ::  BO  :  circumference  great  circle ; 
hence,  surf,  lune  :  surf  sphere  : :  BO  :  cir.  great  circle; 
or,         surf,  lune  :  surf,  sphere  ::  [_BOO :  4  right  angles. 


260  GFOMETRY. 

This  demonstration  assumes  that  BD  is  a  common 
measure  of  the  arc,  BC,  and  the  whole  circumference.  It 
may  happen  that  no  finite  common  measure  can  be 
found ;  but  our  reasoning  would  remain  the  same,  even 
though  this  common  measure  were  to  become  indefinitely 
small. 

Hence  the  proposition. 

Cor.  1.  Any  two  lunes  on  the  same  sphere,  or  on  equal 
spheres,  are  to  each  other  as  their  respective  angles. 

Scholium. —  Spherical  triangles,  formed  by  joining  the  pole  of  an 
arc  of  a  great  circle  with  the  extremities  of  this  arc  by  the  arcs  of 
great  circles,  are  isosceles,  and  contain  two  right  angles.  For  this 
reason  they  are  called  bi-rectangular.  If  the  base  is  also  a  quadrant, 
the  vertex  of  either  angle  becomes  the  pole  of  the  opposite  side,  and 
each  angle  is  measured  by  its  opposite  side.  The  three  angles  are  then 
right  angles,  and  the  triangle  is  for  this  reason  called  tri-rectangular. 
It  is  evident  that  the  surface  of  a  sphere  contains  eight  of  its  tri- 
rectangular  triangles. 

Cor.  2.   Taking  the  right  angle  as  the  unit  of  angles, 
and  denoting  the  angle  of  a  lune  by  A,  and  the  surface 
of  a  tri-rectangular  triangle  by  T,  we  have, 
surf,  of  lune  :  8^  ::  A  :  4; 

whence,       surf,  of  lune  =  2 A  X  T. 

Cor.  3.  A  spherical  ungula  bears  the  same  relation  to 
the  entire  sphere,  that  the  lune,  which  is  the  base  of  the 
ungula,  bears  to  the  surface  of  the  sphere ;  and  hence, 
any  two  spherical  ungulas  in  the  same  sphere,  or  in 
equal  spheres,  are  to  each  other  as  the  angles  of  their  re- 
spective lunes. 

PKOPOSITION    XVI. 

The  area  cf  a  spherical  triangle  is  measured  by  the  excess 
of  the  sum  of  its  angles  over  two  right  angles,  multiplied  by 
the  tri-rectangular  triangle. 

Let  ABC  be  a  spherical  triangle,  and  DJEFLK  the  cir- 
cumference of  the  base  of  the  hemisphere  on  which  this 
triangle  is  situated. 


BOOK    IX.  2G1 

Produce  the  sides  of  the  tri- 
angle until  they  meet  this  cir- 
cumference in  the  points,  D,  U, 
F,  L,  K,  and  P,  thus  forming 
the  sets  of  opposite  triangles, 
DAE,  AKL ;  BEF,  BPK;  CFL, 
CDP. 

Now,  the  triangles  of  each  of 
these  sets  are  together  equal  to 
a  lune,  whose  angle  is  the  cor- 
responding angle  of  the  triangle,  (Prop.  14) ;  hence  we 
have, 

ADAE  +  AAKL  =  2 A  x  T%  (Prop.  15,  Cor.  2). 
ABEF  +  ABPK  =  2B  x  T. 
A  CFL  +  A  CDP  =  2(7  x  T. 

If  the  first  members  of  these  equations  be  added,  it  is 
evident  that  their  sum  will  exceed  the  surface  of  the 
hemisphere  by  twice  the  triangle  ABC;  hence,  adding 
these  equations  member  to  member,  and  substituting  for 
the  first  member  of  the  result  its  value,  4T  +  2  A  ABC, 
we  have 

4I7  +  2aABC  =  2A.T  +  2B.T+  2CT 

or,      2T+    AABC=>   A.T  +    B.T  +    C.T 

whence,  A  ABC =    A.T  +    B.T  +    C.T—2T. 

That  is,      AABC  -  (A  +  B  +  C—  2)  T. 

But  A  +  B  +  C  —  2  is  the  excess  of  the  sum  of  the 
angles  of  the  triangle  over  two  right  angles,  and  T  de- 
notes the  area  of  a  tri-rectangular  triangle. 

Hence  the  proposition  ;  the  area,  etc. 


262  GEOMETRY. 

PROPOSITION    XVII. 

77ie  area  of  any  spherical  polygon  is  measured  by  the  excess 
of  the  sum  of  all  its  angles  over  two  right  angles,  taken  as 
many  times,  less  two,  as  the  polygon  has  sides,  multiplied  by 
the  tri-rectangular  triangle. 

Let  AB  CDE  be  a  spherical  poly-  ^-—"7 

gon;  then  will  its  area  be  meas-  b^"^  / 

ured  by  the  excess  of  the  sum  of  /\  / 

/    \  / 

the  angles,  A,  B,  0,  D,  and  E,  over  / 

two  right  angles  taken  a  number         /  \ 

of  times  which  is  two  less  than     J -~J*Je 

the  number  of  sides,  multiplied  by        \  / 

T,   the    tri-rectangular    triangle.  \.     / 

Through  the  vertex  of  any  of  the  p 

angles,  as  E,  and  the  vertices  of 

the  opposite  angles,  pass  arcs  of  great  circles,  thus  divi- 
ding the  polygon  into  as  many  triangles,  less  two,  as  the 
polygon  has  sides.  The  sum  of  the  angles  of  the  several 
triangles  will  be  equal  to  the  sum  of  the  angles  of  the 
polygon. 

Now,  the  area  of  each  triangle  is  measured  by  the 
excess  of  the  sum  of  its  angles  over  two  right  angles, 
multiplied  by  the  tri-rectangular  triangle.  Hence  the 
sum  of  the  areas  of  all  the  triangles,  or  the  area  of  the 
polygon,  is  measured  by  the  excess  of  the  sum  of  all  the 
angles  of  the  triangles  over  two  right  angles,  taken  as 
many  times  as  there  are  triangles,  multiplied  by  the  tri- 
rectangular  triangle.  But  there  are  as  many  triangles  as 
the  polygon  has  sides,  less  two. 

Hence  the  proposition ;  the  area  of  any  spherical  voly- 
gon,  etc. 

Cor.  If  S  denote  the  sum  of  the  angles  of  any  spherical 
polygon,  n  the  number  of  sides,  and  T  the  tri-rectan- 
gular triangle,  the  right  angle  being  the  unit  of  angles ; 
the  area  of  the  polygon  will  be  expressed  by 

[£_  2  (n  —  2)]  x  T=  (#—  2n  +  4)  T. 


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Standard  French  Course. 

By  Louis  Fasquelle,  LL.  D.,  late  Professor  of  Modern  Languages  in  the  University 

of  Michigan 
The  plan  of  this  popular  Series  embraces  a  combination  of  two  rival  systems:  the 
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principles  and  ordinary  applications  of  Common  Arithmetic,  to  be  used  both  as  a  class- 
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The  design  is  to  furnish  a  large  number  of  well-prepared  Intellectual  and  Written 
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There  is  a  general  classification  of  the  examples,  the  work  being  divided  into  six 
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Two  editions  are  printed,  one  for  the  use  of  Teachers,  with  answers ;  the  other,  without 
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From  Ohio  Educational  Journal,  E.  E.  White  ( 'author  of 'White 's  Arithmetics ),  Editor : 
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"First  Steps  in  Music,  by  Prof.  George  B.  Loomis,  seems  tome  admirably  adapted 

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From  G.  A.  Chase,  Prin.  Louisville,  Ky.y  Female  High  School. 
"  I  have  tried  Mr.  Loomis1  Plan  with  the  little  pupils  in  the  school  of  a  friend  of  mine. 
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"  I  knew  of  no  other  attempt  {' Loomis1  First  Steps')  so  successful  to  bring  the  elemen- 
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1  i 


